Post on 21-Dec-2015
12INFINITE SEQUENCES AND SERIESINFINITE SEQUENCES AND SERIES
Infinite sequences and series were
introduced briefly in A Preview of Calculus
in connection with Zenorsquos paradoxes and
the decimal representation of numbers
INFINITE SEQUENCES AND SERIES
Their importance in calculus stems from
Newtonrsquos idea of representing functions as
sums of infinite series
For instance in finding areas he often integrated a function by first expressing it as a series and then integrating each term of the series
INFINITE SEQUENCES AND SERIES
We will pursue his idea in Section 1210
in order to integrate such functions as e-x2
Recall that we have previously been unable to do this
INFINITE SEQUENCES AND SERIES
Many of the functions that arise in
mathematical physics and chemistry
such as Bessel functions are defined
as sums of series
It is important to be familiar with the basic concepts of convergence of infinite sequences and series
INFINITE SEQUENCES AND SERIES
Physicists also use series in another
way as we will see in Section 1211
In studying fields as diverse as optics special relativity and electromagnetism they analyze phenomena by replacing a function with the first few terms in the series that represents it
INFINITE SEQUENCES AND SERIES
121Sequences
In this section we will learn about
Various concepts related to sequences
INFINITE SEQUENCES AND SERIES
SEQUENCE
A sequence can be thought of as a list
of numbers written in a definite order
a1 a2 a3 a4 hellip an hellip
The number a1 is called the first term a2 is the second term and in general an is the nth term
SEQUENCES
We will deal exclusively with infinite
sequences
So each term an will have a successor an+1
SEQUENCES
Notice that for every positive integer n
there is a corresponding number an
So a sequence can be defined as
A function whose domain is the set of positive integers
SEQUENCES
However we usually write an instead
of the function notation f(n) for the value
of the function at the number n
SEQUENCES
The sequence a1 a2 a3 is also
denoted by
1orn n n
a a
Notation
SEQUENCES
Some sequences can be
defined by giving a formula
for the nth term
Example 1
SEQUENCES
In the following examples we give three
descriptions of the sequence
1 Using the preceding notation
2 Using the defining formula
3 Writing out the terms of the sequence
Example 1
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding
Notation
Defining
Formula
Terms of
Sequence
11 n
n
n
1n
na
n
1 2 3 4
2 3 4 5 1
n
n
SEQUENCES Example 1 b
Preceding
Notation
Defining
Formula
Terms of
Sequence
( 1) ( 1)
3
n
n
n
( 1) ( 1)
3
n
n n
na
2 3 4 5
3 9 27 81
( 1) ( 1)
3
n
n
n
SEQUENCES Example 1 c
Preceding
Notation
Defining
Formula
Terms of
Sequence
3
3n
n
3
( 3)na n
n
01 2 3 3n
SEQUENCES Example 1 d
Preceding
Notation
Defining
Formula
Terms of
Sequence
0
cos6 n
n
cos6
( 0)
n
na
n
3 11 0 cos
2 2 6
n
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms
continues
3 4 5 6 7
5 25 125 625 3125
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 5
5 25 125
6 7
625 3125
a a a
a a
Example 2
SEQUENCES
Notice that the numerators of these fractions
start with 3 and increase by 1 whenever we
go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers
of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately
positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant
we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)
5n
n n
na
Example 2
SEQUENCES
We now look at some sequences
that donrsquot have a simple defining
equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as
of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place
of the number e then an is a well-defined
sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined
recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
Infinite sequences and series were
introduced briefly in A Preview of Calculus
in connection with Zenorsquos paradoxes and
the decimal representation of numbers
INFINITE SEQUENCES AND SERIES
Their importance in calculus stems from
Newtonrsquos idea of representing functions as
sums of infinite series
For instance in finding areas he often integrated a function by first expressing it as a series and then integrating each term of the series
INFINITE SEQUENCES AND SERIES
We will pursue his idea in Section 1210
in order to integrate such functions as e-x2
Recall that we have previously been unable to do this
INFINITE SEQUENCES AND SERIES
Many of the functions that arise in
mathematical physics and chemistry
such as Bessel functions are defined
as sums of series
It is important to be familiar with the basic concepts of convergence of infinite sequences and series
INFINITE SEQUENCES AND SERIES
Physicists also use series in another
way as we will see in Section 1211
In studying fields as diverse as optics special relativity and electromagnetism they analyze phenomena by replacing a function with the first few terms in the series that represents it
INFINITE SEQUENCES AND SERIES
121Sequences
In this section we will learn about
Various concepts related to sequences
INFINITE SEQUENCES AND SERIES
SEQUENCE
A sequence can be thought of as a list
of numbers written in a definite order
a1 a2 a3 a4 hellip an hellip
The number a1 is called the first term a2 is the second term and in general an is the nth term
SEQUENCES
We will deal exclusively with infinite
sequences
So each term an will have a successor an+1
SEQUENCES
Notice that for every positive integer n
there is a corresponding number an
So a sequence can be defined as
A function whose domain is the set of positive integers
SEQUENCES
However we usually write an instead
of the function notation f(n) for the value
of the function at the number n
SEQUENCES
The sequence a1 a2 a3 is also
denoted by
1orn n n
a a
Notation
SEQUENCES
Some sequences can be
defined by giving a formula
for the nth term
Example 1
SEQUENCES
In the following examples we give three
descriptions of the sequence
1 Using the preceding notation
2 Using the defining formula
3 Writing out the terms of the sequence
Example 1
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding
Notation
Defining
Formula
Terms of
Sequence
11 n
n
n
1n
na
n
1 2 3 4
2 3 4 5 1
n
n
SEQUENCES Example 1 b
Preceding
Notation
Defining
Formula
Terms of
Sequence
( 1) ( 1)
3
n
n
n
( 1) ( 1)
3
n
n n
na
2 3 4 5
3 9 27 81
( 1) ( 1)
3
n
n
n
SEQUENCES Example 1 c
Preceding
Notation
Defining
Formula
Terms of
Sequence
3
3n
n
3
( 3)na n
n
01 2 3 3n
SEQUENCES Example 1 d
Preceding
Notation
Defining
Formula
Terms of
Sequence
0
cos6 n
n
cos6
( 0)
n
na
n
3 11 0 cos
2 2 6
n
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms
continues
3 4 5 6 7
5 25 125 625 3125
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 5
5 25 125
6 7
625 3125
a a a
a a
Example 2
SEQUENCES
Notice that the numerators of these fractions
start with 3 and increase by 1 whenever we
go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers
of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately
positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant
we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)
5n
n n
na
Example 2
SEQUENCES
We now look at some sequences
that donrsquot have a simple defining
equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as
of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place
of the number e then an is a well-defined
sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined
recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
Their importance in calculus stems from
Newtonrsquos idea of representing functions as
sums of infinite series
For instance in finding areas he often integrated a function by first expressing it as a series and then integrating each term of the series
INFINITE SEQUENCES AND SERIES
We will pursue his idea in Section 1210
in order to integrate such functions as e-x2
Recall that we have previously been unable to do this
INFINITE SEQUENCES AND SERIES
Many of the functions that arise in
mathematical physics and chemistry
such as Bessel functions are defined
as sums of series
It is important to be familiar with the basic concepts of convergence of infinite sequences and series
INFINITE SEQUENCES AND SERIES
Physicists also use series in another
way as we will see in Section 1211
In studying fields as diverse as optics special relativity and electromagnetism they analyze phenomena by replacing a function with the first few terms in the series that represents it
INFINITE SEQUENCES AND SERIES
121Sequences
In this section we will learn about
Various concepts related to sequences
INFINITE SEQUENCES AND SERIES
SEQUENCE
A sequence can be thought of as a list
of numbers written in a definite order
a1 a2 a3 a4 hellip an hellip
The number a1 is called the first term a2 is the second term and in general an is the nth term
SEQUENCES
We will deal exclusively with infinite
sequences
So each term an will have a successor an+1
SEQUENCES
Notice that for every positive integer n
there is a corresponding number an
So a sequence can be defined as
A function whose domain is the set of positive integers
SEQUENCES
However we usually write an instead
of the function notation f(n) for the value
of the function at the number n
SEQUENCES
The sequence a1 a2 a3 is also
denoted by
1orn n n
a a
Notation
SEQUENCES
Some sequences can be
defined by giving a formula
for the nth term
Example 1
SEQUENCES
In the following examples we give three
descriptions of the sequence
1 Using the preceding notation
2 Using the defining formula
3 Writing out the terms of the sequence
Example 1
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding
Notation
Defining
Formula
Terms of
Sequence
11 n
n
n
1n
na
n
1 2 3 4
2 3 4 5 1
n
n
SEQUENCES Example 1 b
Preceding
Notation
Defining
Formula
Terms of
Sequence
( 1) ( 1)
3
n
n
n
( 1) ( 1)
3
n
n n
na
2 3 4 5
3 9 27 81
( 1) ( 1)
3
n
n
n
SEQUENCES Example 1 c
Preceding
Notation
Defining
Formula
Terms of
Sequence
3
3n
n
3
( 3)na n
n
01 2 3 3n
SEQUENCES Example 1 d
Preceding
Notation
Defining
Formula
Terms of
Sequence
0
cos6 n
n
cos6
( 0)
n
na
n
3 11 0 cos
2 2 6
n
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms
continues
3 4 5 6 7
5 25 125 625 3125
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 5
5 25 125
6 7
625 3125
a a a
a a
Example 2
SEQUENCES
Notice that the numerators of these fractions
start with 3 and increase by 1 whenever we
go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers
of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately
positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant
we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)
5n
n n
na
Example 2
SEQUENCES
We now look at some sequences
that donrsquot have a simple defining
equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as
of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place
of the number e then an is a well-defined
sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined
recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
We will pursue his idea in Section 1210
in order to integrate such functions as e-x2
Recall that we have previously been unable to do this
INFINITE SEQUENCES AND SERIES
Many of the functions that arise in
mathematical physics and chemistry
such as Bessel functions are defined
as sums of series
It is important to be familiar with the basic concepts of convergence of infinite sequences and series
INFINITE SEQUENCES AND SERIES
Physicists also use series in another
way as we will see in Section 1211
In studying fields as diverse as optics special relativity and electromagnetism they analyze phenomena by replacing a function with the first few terms in the series that represents it
INFINITE SEQUENCES AND SERIES
121Sequences
In this section we will learn about
Various concepts related to sequences
INFINITE SEQUENCES AND SERIES
SEQUENCE
A sequence can be thought of as a list
of numbers written in a definite order
a1 a2 a3 a4 hellip an hellip
The number a1 is called the first term a2 is the second term and in general an is the nth term
SEQUENCES
We will deal exclusively with infinite
sequences
So each term an will have a successor an+1
SEQUENCES
Notice that for every positive integer n
there is a corresponding number an
So a sequence can be defined as
A function whose domain is the set of positive integers
SEQUENCES
However we usually write an instead
of the function notation f(n) for the value
of the function at the number n
SEQUENCES
The sequence a1 a2 a3 is also
denoted by
1orn n n
a a
Notation
SEQUENCES
Some sequences can be
defined by giving a formula
for the nth term
Example 1
SEQUENCES
In the following examples we give three
descriptions of the sequence
1 Using the preceding notation
2 Using the defining formula
3 Writing out the terms of the sequence
Example 1
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding
Notation
Defining
Formula
Terms of
Sequence
11 n
n
n
1n
na
n
1 2 3 4
2 3 4 5 1
n
n
SEQUENCES Example 1 b
Preceding
Notation
Defining
Formula
Terms of
Sequence
( 1) ( 1)
3
n
n
n
( 1) ( 1)
3
n
n n
na
2 3 4 5
3 9 27 81
( 1) ( 1)
3
n
n
n
SEQUENCES Example 1 c
Preceding
Notation
Defining
Formula
Terms of
Sequence
3
3n
n
3
( 3)na n
n
01 2 3 3n
SEQUENCES Example 1 d
Preceding
Notation
Defining
Formula
Terms of
Sequence
0
cos6 n
n
cos6
( 0)
n
na
n
3 11 0 cos
2 2 6
n
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms
continues
3 4 5 6 7
5 25 125 625 3125
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 5
5 25 125
6 7
625 3125
a a a
a a
Example 2
SEQUENCES
Notice that the numerators of these fractions
start with 3 and increase by 1 whenever we
go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers
of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately
positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant
we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)
5n
n n
na
Example 2
SEQUENCES
We now look at some sequences
that donrsquot have a simple defining
equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as
of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place
of the number e then an is a well-defined
sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined
recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
Many of the functions that arise in
mathematical physics and chemistry
such as Bessel functions are defined
as sums of series
It is important to be familiar with the basic concepts of convergence of infinite sequences and series
INFINITE SEQUENCES AND SERIES
Physicists also use series in another
way as we will see in Section 1211
In studying fields as diverse as optics special relativity and electromagnetism they analyze phenomena by replacing a function with the first few terms in the series that represents it
INFINITE SEQUENCES AND SERIES
121Sequences
In this section we will learn about
Various concepts related to sequences
INFINITE SEQUENCES AND SERIES
SEQUENCE
A sequence can be thought of as a list
of numbers written in a definite order
a1 a2 a3 a4 hellip an hellip
The number a1 is called the first term a2 is the second term and in general an is the nth term
SEQUENCES
We will deal exclusively with infinite
sequences
So each term an will have a successor an+1
SEQUENCES
Notice that for every positive integer n
there is a corresponding number an
So a sequence can be defined as
A function whose domain is the set of positive integers
SEQUENCES
However we usually write an instead
of the function notation f(n) for the value
of the function at the number n
SEQUENCES
The sequence a1 a2 a3 is also
denoted by
1orn n n
a a
Notation
SEQUENCES
Some sequences can be
defined by giving a formula
for the nth term
Example 1
SEQUENCES
In the following examples we give three
descriptions of the sequence
1 Using the preceding notation
2 Using the defining formula
3 Writing out the terms of the sequence
Example 1
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding
Notation
Defining
Formula
Terms of
Sequence
11 n
n
n
1n
na
n
1 2 3 4
2 3 4 5 1
n
n
SEQUENCES Example 1 b
Preceding
Notation
Defining
Formula
Terms of
Sequence
( 1) ( 1)
3
n
n
n
( 1) ( 1)
3
n
n n
na
2 3 4 5
3 9 27 81
( 1) ( 1)
3
n
n
n
SEQUENCES Example 1 c
Preceding
Notation
Defining
Formula
Terms of
Sequence
3
3n
n
3
( 3)na n
n
01 2 3 3n
SEQUENCES Example 1 d
Preceding
Notation
Defining
Formula
Terms of
Sequence
0
cos6 n
n
cos6
( 0)
n
na
n
3 11 0 cos
2 2 6
n
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms
continues
3 4 5 6 7
5 25 125 625 3125
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 5
5 25 125
6 7
625 3125
a a a
a a
Example 2
SEQUENCES
Notice that the numerators of these fractions
start with 3 and increase by 1 whenever we
go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers
of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately
positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant
we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)
5n
n n
na
Example 2
SEQUENCES
We now look at some sequences
that donrsquot have a simple defining
equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as
of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place
of the number e then an is a well-defined
sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined
recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
Physicists also use series in another
way as we will see in Section 1211
In studying fields as diverse as optics special relativity and electromagnetism they analyze phenomena by replacing a function with the first few terms in the series that represents it
INFINITE SEQUENCES AND SERIES
121Sequences
In this section we will learn about
Various concepts related to sequences
INFINITE SEQUENCES AND SERIES
SEQUENCE
A sequence can be thought of as a list
of numbers written in a definite order
a1 a2 a3 a4 hellip an hellip
The number a1 is called the first term a2 is the second term and in general an is the nth term
SEQUENCES
We will deal exclusively with infinite
sequences
So each term an will have a successor an+1
SEQUENCES
Notice that for every positive integer n
there is a corresponding number an
So a sequence can be defined as
A function whose domain is the set of positive integers
SEQUENCES
However we usually write an instead
of the function notation f(n) for the value
of the function at the number n
SEQUENCES
The sequence a1 a2 a3 is also
denoted by
1orn n n
a a
Notation
SEQUENCES
Some sequences can be
defined by giving a formula
for the nth term
Example 1
SEQUENCES
In the following examples we give three
descriptions of the sequence
1 Using the preceding notation
2 Using the defining formula
3 Writing out the terms of the sequence
Example 1
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding
Notation
Defining
Formula
Terms of
Sequence
11 n
n
n
1n
na
n
1 2 3 4
2 3 4 5 1
n
n
SEQUENCES Example 1 b
Preceding
Notation
Defining
Formula
Terms of
Sequence
( 1) ( 1)
3
n
n
n
( 1) ( 1)
3
n
n n
na
2 3 4 5
3 9 27 81
( 1) ( 1)
3
n
n
n
SEQUENCES Example 1 c
Preceding
Notation
Defining
Formula
Terms of
Sequence
3
3n
n
3
( 3)na n
n
01 2 3 3n
SEQUENCES Example 1 d
Preceding
Notation
Defining
Formula
Terms of
Sequence
0
cos6 n
n
cos6
( 0)
n
na
n
3 11 0 cos
2 2 6
n
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms
continues
3 4 5 6 7
5 25 125 625 3125
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 5
5 25 125
6 7
625 3125
a a a
a a
Example 2
SEQUENCES
Notice that the numerators of these fractions
start with 3 and increase by 1 whenever we
go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers
of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately
positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant
we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)
5n
n n
na
Example 2
SEQUENCES
We now look at some sequences
that donrsquot have a simple defining
equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as
of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place
of the number e then an is a well-defined
sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined
recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
121Sequences
In this section we will learn about
Various concepts related to sequences
INFINITE SEQUENCES AND SERIES
SEQUENCE
A sequence can be thought of as a list
of numbers written in a definite order
a1 a2 a3 a4 hellip an hellip
The number a1 is called the first term a2 is the second term and in general an is the nth term
SEQUENCES
We will deal exclusively with infinite
sequences
So each term an will have a successor an+1
SEQUENCES
Notice that for every positive integer n
there is a corresponding number an
So a sequence can be defined as
A function whose domain is the set of positive integers
SEQUENCES
However we usually write an instead
of the function notation f(n) for the value
of the function at the number n
SEQUENCES
The sequence a1 a2 a3 is also
denoted by
1orn n n
a a
Notation
SEQUENCES
Some sequences can be
defined by giving a formula
for the nth term
Example 1
SEQUENCES
In the following examples we give three
descriptions of the sequence
1 Using the preceding notation
2 Using the defining formula
3 Writing out the terms of the sequence
Example 1
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding
Notation
Defining
Formula
Terms of
Sequence
11 n
n
n
1n
na
n
1 2 3 4
2 3 4 5 1
n
n
SEQUENCES Example 1 b
Preceding
Notation
Defining
Formula
Terms of
Sequence
( 1) ( 1)
3
n
n
n
( 1) ( 1)
3
n
n n
na
2 3 4 5
3 9 27 81
( 1) ( 1)
3
n
n
n
SEQUENCES Example 1 c
Preceding
Notation
Defining
Formula
Terms of
Sequence
3
3n
n
3
( 3)na n
n
01 2 3 3n
SEQUENCES Example 1 d
Preceding
Notation
Defining
Formula
Terms of
Sequence
0
cos6 n
n
cos6
( 0)
n
na
n
3 11 0 cos
2 2 6
n
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms
continues
3 4 5 6 7
5 25 125 625 3125
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 5
5 25 125
6 7
625 3125
a a a
a a
Example 2
SEQUENCES
Notice that the numerators of these fractions
start with 3 and increase by 1 whenever we
go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers
of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately
positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant
we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)
5n
n n
na
Example 2
SEQUENCES
We now look at some sequences
that donrsquot have a simple defining
equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as
of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place
of the number e then an is a well-defined
sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined
recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
SEQUENCE
A sequence can be thought of as a list
of numbers written in a definite order
a1 a2 a3 a4 hellip an hellip
The number a1 is called the first term a2 is the second term and in general an is the nth term
SEQUENCES
We will deal exclusively with infinite
sequences
So each term an will have a successor an+1
SEQUENCES
Notice that for every positive integer n
there is a corresponding number an
So a sequence can be defined as
A function whose domain is the set of positive integers
SEQUENCES
However we usually write an instead
of the function notation f(n) for the value
of the function at the number n
SEQUENCES
The sequence a1 a2 a3 is also
denoted by
1orn n n
a a
Notation
SEQUENCES
Some sequences can be
defined by giving a formula
for the nth term
Example 1
SEQUENCES
In the following examples we give three
descriptions of the sequence
1 Using the preceding notation
2 Using the defining formula
3 Writing out the terms of the sequence
Example 1
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding
Notation
Defining
Formula
Terms of
Sequence
11 n
n
n
1n
na
n
1 2 3 4
2 3 4 5 1
n
n
SEQUENCES Example 1 b
Preceding
Notation
Defining
Formula
Terms of
Sequence
( 1) ( 1)
3
n
n
n
( 1) ( 1)
3
n
n n
na
2 3 4 5
3 9 27 81
( 1) ( 1)
3
n
n
n
SEQUENCES Example 1 c
Preceding
Notation
Defining
Formula
Terms of
Sequence
3
3n
n
3
( 3)na n
n
01 2 3 3n
SEQUENCES Example 1 d
Preceding
Notation
Defining
Formula
Terms of
Sequence
0
cos6 n
n
cos6
( 0)
n
na
n
3 11 0 cos
2 2 6
n
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms
continues
3 4 5 6 7
5 25 125 625 3125
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 5
5 25 125
6 7
625 3125
a a a
a a
Example 2
SEQUENCES
Notice that the numerators of these fractions
start with 3 and increase by 1 whenever we
go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers
of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately
positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant
we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)
5n
n n
na
Example 2
SEQUENCES
We now look at some sequences
that donrsquot have a simple defining
equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as
of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place
of the number e then an is a well-defined
sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined
recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
SEQUENCES
We will deal exclusively with infinite
sequences
So each term an will have a successor an+1
SEQUENCES
Notice that for every positive integer n
there is a corresponding number an
So a sequence can be defined as
A function whose domain is the set of positive integers
SEQUENCES
However we usually write an instead
of the function notation f(n) for the value
of the function at the number n
SEQUENCES
The sequence a1 a2 a3 is also
denoted by
1orn n n
a a
Notation
SEQUENCES
Some sequences can be
defined by giving a formula
for the nth term
Example 1
SEQUENCES
In the following examples we give three
descriptions of the sequence
1 Using the preceding notation
2 Using the defining formula
3 Writing out the terms of the sequence
Example 1
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding
Notation
Defining
Formula
Terms of
Sequence
11 n
n
n
1n
na
n
1 2 3 4
2 3 4 5 1
n
n
SEQUENCES Example 1 b
Preceding
Notation
Defining
Formula
Terms of
Sequence
( 1) ( 1)
3
n
n
n
( 1) ( 1)
3
n
n n
na
2 3 4 5
3 9 27 81
( 1) ( 1)
3
n
n
n
SEQUENCES Example 1 c
Preceding
Notation
Defining
Formula
Terms of
Sequence
3
3n
n
3
( 3)na n
n
01 2 3 3n
SEQUENCES Example 1 d
Preceding
Notation
Defining
Formula
Terms of
Sequence
0
cos6 n
n
cos6
( 0)
n
na
n
3 11 0 cos
2 2 6
n
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms
continues
3 4 5 6 7
5 25 125 625 3125
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 5
5 25 125
6 7
625 3125
a a a
a a
Example 2
SEQUENCES
Notice that the numerators of these fractions
start with 3 and increase by 1 whenever we
go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers
of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately
positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant
we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)
5n
n n
na
Example 2
SEQUENCES
We now look at some sequences
that donrsquot have a simple defining
equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as
of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place
of the number e then an is a well-defined
sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined
recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
SEQUENCES
Notice that for every positive integer n
there is a corresponding number an
So a sequence can be defined as
A function whose domain is the set of positive integers
SEQUENCES
However we usually write an instead
of the function notation f(n) for the value
of the function at the number n
SEQUENCES
The sequence a1 a2 a3 is also
denoted by
1orn n n
a a
Notation
SEQUENCES
Some sequences can be
defined by giving a formula
for the nth term
Example 1
SEQUENCES
In the following examples we give three
descriptions of the sequence
1 Using the preceding notation
2 Using the defining formula
3 Writing out the terms of the sequence
Example 1
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding
Notation
Defining
Formula
Terms of
Sequence
11 n
n
n
1n
na
n
1 2 3 4
2 3 4 5 1
n
n
SEQUENCES Example 1 b
Preceding
Notation
Defining
Formula
Terms of
Sequence
( 1) ( 1)
3
n
n
n
( 1) ( 1)
3
n
n n
na
2 3 4 5
3 9 27 81
( 1) ( 1)
3
n
n
n
SEQUENCES Example 1 c
Preceding
Notation
Defining
Formula
Terms of
Sequence
3
3n
n
3
( 3)na n
n
01 2 3 3n
SEQUENCES Example 1 d
Preceding
Notation
Defining
Formula
Terms of
Sequence
0
cos6 n
n
cos6
( 0)
n
na
n
3 11 0 cos
2 2 6
n
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms
continues
3 4 5 6 7
5 25 125 625 3125
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 5
5 25 125
6 7
625 3125
a a a
a a
Example 2
SEQUENCES
Notice that the numerators of these fractions
start with 3 and increase by 1 whenever we
go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers
of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately
positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant
we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)
5n
n n
na
Example 2
SEQUENCES
We now look at some sequences
that donrsquot have a simple defining
equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as
of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place
of the number e then an is a well-defined
sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined
recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
SEQUENCES
However we usually write an instead
of the function notation f(n) for the value
of the function at the number n
SEQUENCES
The sequence a1 a2 a3 is also
denoted by
1orn n n
a a
Notation
SEQUENCES
Some sequences can be
defined by giving a formula
for the nth term
Example 1
SEQUENCES
In the following examples we give three
descriptions of the sequence
1 Using the preceding notation
2 Using the defining formula
3 Writing out the terms of the sequence
Example 1
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding
Notation
Defining
Formula
Terms of
Sequence
11 n
n
n
1n
na
n
1 2 3 4
2 3 4 5 1
n
n
SEQUENCES Example 1 b
Preceding
Notation
Defining
Formula
Terms of
Sequence
( 1) ( 1)
3
n
n
n
( 1) ( 1)
3
n
n n
na
2 3 4 5
3 9 27 81
( 1) ( 1)
3
n
n
n
SEQUENCES Example 1 c
Preceding
Notation
Defining
Formula
Terms of
Sequence
3
3n
n
3
( 3)na n
n
01 2 3 3n
SEQUENCES Example 1 d
Preceding
Notation
Defining
Formula
Terms of
Sequence
0
cos6 n
n
cos6
( 0)
n
na
n
3 11 0 cos
2 2 6
n
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms
continues
3 4 5 6 7
5 25 125 625 3125
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 5
5 25 125
6 7
625 3125
a a a
a a
Example 2
SEQUENCES
Notice that the numerators of these fractions
start with 3 and increase by 1 whenever we
go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers
of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately
positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant
we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)
5n
n n
na
Example 2
SEQUENCES
We now look at some sequences
that donrsquot have a simple defining
equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as
of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place
of the number e then an is a well-defined
sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined
recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
SEQUENCES
The sequence a1 a2 a3 is also
denoted by
1orn n n
a a
Notation
SEQUENCES
Some sequences can be
defined by giving a formula
for the nth term
Example 1
SEQUENCES
In the following examples we give three
descriptions of the sequence
1 Using the preceding notation
2 Using the defining formula
3 Writing out the terms of the sequence
Example 1
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding
Notation
Defining
Formula
Terms of
Sequence
11 n
n
n
1n
na
n
1 2 3 4
2 3 4 5 1
n
n
SEQUENCES Example 1 b
Preceding
Notation
Defining
Formula
Terms of
Sequence
( 1) ( 1)
3
n
n
n
( 1) ( 1)
3
n
n n
na
2 3 4 5
3 9 27 81
( 1) ( 1)
3
n
n
n
SEQUENCES Example 1 c
Preceding
Notation
Defining
Formula
Terms of
Sequence
3
3n
n
3
( 3)na n
n
01 2 3 3n
SEQUENCES Example 1 d
Preceding
Notation
Defining
Formula
Terms of
Sequence
0
cos6 n
n
cos6
( 0)
n
na
n
3 11 0 cos
2 2 6
n
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms
continues
3 4 5 6 7
5 25 125 625 3125
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 5
5 25 125
6 7
625 3125
a a a
a a
Example 2
SEQUENCES
Notice that the numerators of these fractions
start with 3 and increase by 1 whenever we
go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers
of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately
positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant
we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)
5n
n n
na
Example 2
SEQUENCES
We now look at some sequences
that donrsquot have a simple defining
equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as
of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place
of the number e then an is a well-defined
sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined
recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
SEQUENCES
Some sequences can be
defined by giving a formula
for the nth term
Example 1
SEQUENCES
In the following examples we give three
descriptions of the sequence
1 Using the preceding notation
2 Using the defining formula
3 Writing out the terms of the sequence
Example 1
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding
Notation
Defining
Formula
Terms of
Sequence
11 n
n
n
1n
na
n
1 2 3 4
2 3 4 5 1
n
n
SEQUENCES Example 1 b
Preceding
Notation
Defining
Formula
Terms of
Sequence
( 1) ( 1)
3
n
n
n
( 1) ( 1)
3
n
n n
na
2 3 4 5
3 9 27 81
( 1) ( 1)
3
n
n
n
SEQUENCES Example 1 c
Preceding
Notation
Defining
Formula
Terms of
Sequence
3
3n
n
3
( 3)na n
n
01 2 3 3n
SEQUENCES Example 1 d
Preceding
Notation
Defining
Formula
Terms of
Sequence
0
cos6 n
n
cos6
( 0)
n
na
n
3 11 0 cos
2 2 6
n
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms
continues
3 4 5 6 7
5 25 125 625 3125
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 5
5 25 125
6 7
625 3125
a a a
a a
Example 2
SEQUENCES
Notice that the numerators of these fractions
start with 3 and increase by 1 whenever we
go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers
of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately
positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant
we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)
5n
n n
na
Example 2
SEQUENCES
We now look at some sequences
that donrsquot have a simple defining
equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as
of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place
of the number e then an is a well-defined
sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined
recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
SEQUENCES
In the following examples we give three
descriptions of the sequence
1 Using the preceding notation
2 Using the defining formula
3 Writing out the terms of the sequence
Example 1
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding
Notation
Defining
Formula
Terms of
Sequence
11 n
n
n
1n
na
n
1 2 3 4
2 3 4 5 1
n
n
SEQUENCES Example 1 b
Preceding
Notation
Defining
Formula
Terms of
Sequence
( 1) ( 1)
3
n
n
n
( 1) ( 1)
3
n
n n
na
2 3 4 5
3 9 27 81
( 1) ( 1)
3
n
n
n
SEQUENCES Example 1 c
Preceding
Notation
Defining
Formula
Terms of
Sequence
3
3n
n
3
( 3)na n
n
01 2 3 3n
SEQUENCES Example 1 d
Preceding
Notation
Defining
Formula
Terms of
Sequence
0
cos6 n
n
cos6
( 0)
n
na
n
3 11 0 cos
2 2 6
n
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms
continues
3 4 5 6 7
5 25 125 625 3125
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 5
5 25 125
6 7
625 3125
a a a
a a
Example 2
SEQUENCES
Notice that the numerators of these fractions
start with 3 and increase by 1 whenever we
go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers
of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately
positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant
we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)
5n
n n
na
Example 2
SEQUENCES
We now look at some sequences
that donrsquot have a simple defining
equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as
of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place
of the number e then an is a well-defined
sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined
recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding
Notation
Defining
Formula
Terms of
Sequence
11 n
n
n
1n
na
n
1 2 3 4
2 3 4 5 1
n
n
SEQUENCES Example 1 b
Preceding
Notation
Defining
Formula
Terms of
Sequence
( 1) ( 1)
3
n
n
n
( 1) ( 1)
3
n
n n
na
2 3 4 5
3 9 27 81
( 1) ( 1)
3
n
n
n
SEQUENCES Example 1 c
Preceding
Notation
Defining
Formula
Terms of
Sequence
3
3n
n
3
( 3)na n
n
01 2 3 3n
SEQUENCES Example 1 d
Preceding
Notation
Defining
Formula
Terms of
Sequence
0
cos6 n
n
cos6
( 0)
n
na
n
3 11 0 cos
2 2 6
n
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms
continues
3 4 5 6 7
5 25 125 625 3125
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 5
5 25 125
6 7
625 3125
a a a
a a
Example 2
SEQUENCES
Notice that the numerators of these fractions
start with 3 and increase by 1 whenever we
go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers
of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately
positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant
we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)
5n
n n
na
Example 2
SEQUENCES
We now look at some sequences
that donrsquot have a simple defining
equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as
of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place
of the number e then an is a well-defined
sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined
recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
SEQUENCES Example 1 b
Preceding
Notation
Defining
Formula
Terms of
Sequence
( 1) ( 1)
3
n
n
n
( 1) ( 1)
3
n
n n
na
2 3 4 5
3 9 27 81
( 1) ( 1)
3
n
n
n
SEQUENCES Example 1 c
Preceding
Notation
Defining
Formula
Terms of
Sequence
3
3n
n
3
( 3)na n
n
01 2 3 3n
SEQUENCES Example 1 d
Preceding
Notation
Defining
Formula
Terms of
Sequence
0
cos6 n
n
cos6
( 0)
n
na
n
3 11 0 cos
2 2 6
n
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms
continues
3 4 5 6 7
5 25 125 625 3125
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 5
5 25 125
6 7
625 3125
a a a
a a
Example 2
SEQUENCES
Notice that the numerators of these fractions
start with 3 and increase by 1 whenever we
go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers
of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately
positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant
we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)
5n
n n
na
Example 2
SEQUENCES
We now look at some sequences
that donrsquot have a simple defining
equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as
of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place
of the number e then an is a well-defined
sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined
recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
SEQUENCES Example 1 c
Preceding
Notation
Defining
Formula
Terms of
Sequence
3
3n
n
3
( 3)na n
n
01 2 3 3n
SEQUENCES Example 1 d
Preceding
Notation
Defining
Formula
Terms of
Sequence
0
cos6 n
n
cos6
( 0)
n
na
n
3 11 0 cos
2 2 6
n
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms
continues
3 4 5 6 7
5 25 125 625 3125
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 5
5 25 125
6 7
625 3125
a a a
a a
Example 2
SEQUENCES
Notice that the numerators of these fractions
start with 3 and increase by 1 whenever we
go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers
of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately
positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant
we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)
5n
n n
na
Example 2
SEQUENCES
We now look at some sequences
that donrsquot have a simple defining
equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as
of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place
of the number e then an is a well-defined
sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined
recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
SEQUENCES Example 1 d
Preceding
Notation
Defining
Formula
Terms of
Sequence
0
cos6 n
n
cos6
( 0)
n
na
n
3 11 0 cos
2 2 6
n
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms
continues
3 4 5 6 7
5 25 125 625 3125
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 5
5 25 125
6 7
625 3125
a a a
a a
Example 2
SEQUENCES
Notice that the numerators of these fractions
start with 3 and increase by 1 whenever we
go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers
of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately
positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant
we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)
5n
n n
na
Example 2
SEQUENCES
We now look at some sequences
that donrsquot have a simple defining
equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as
of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place
of the number e then an is a well-defined
sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined
recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms
continues
3 4 5 6 7
5 25 125 625 3125
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 5
5 25 125
6 7
625 3125
a a a
a a
Example 2
SEQUENCES
Notice that the numerators of these fractions
start with 3 and increase by 1 whenever we
go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers
of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately
positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant
we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)
5n
n n
na
Example 2
SEQUENCES
We now look at some sequences
that donrsquot have a simple defining
equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as
of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place
of the number e then an is a well-defined
sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined
recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
SEQUENCES
We are given that
1 2 3
4 5
3 4 5
5 25 125
6 7
625 3125
a a a
a a
Example 2
SEQUENCES
Notice that the numerators of these fractions
start with 3 and increase by 1 whenever we
go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers
of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately
positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant
we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)
5n
n n
na
Example 2
SEQUENCES
We now look at some sequences
that donrsquot have a simple defining
equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as
of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place
of the number e then an is a well-defined
sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined
recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
SEQUENCES
Notice that the numerators of these fractions
start with 3 and increase by 1 whenever we
go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers
of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately
positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant
we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)
5n
n n
na
Example 2
SEQUENCES
We now look at some sequences
that donrsquot have a simple defining
equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as
of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place
of the number e then an is a well-defined
sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined
recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
SEQUENCES
The denominators are the powers
of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately
positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant
we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)
5n
n n
na
Example 2
SEQUENCES
We now look at some sequences
that donrsquot have a simple defining
equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as
of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place
of the number e then an is a well-defined
sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined
recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
SEQUENCES
The signs of the terms are alternately
positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant
we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)
5n
n n
na
Example 2
SEQUENCES
We now look at some sequences
that donrsquot have a simple defining
equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as
of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place
of the number e then an is a well-defined
sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined
recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
SEQUENCES
In Example 1 b the factor (ndash1)n meant
we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)
5n
n n
na
Example 2
SEQUENCES
We now look at some sequences
that donrsquot have a simple defining
equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as
of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place
of the number e then an is a well-defined
sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined
recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)
5n
n n
na
Example 2
SEQUENCES
We now look at some sequences
that donrsquot have a simple defining
equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as
of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place
of the number e then an is a well-defined
sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined
recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
SEQUENCES
We now look at some sequences
that donrsquot have a simple defining
equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as
of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place
of the number e then an is a well-defined
sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined
recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
SEQUENCES
The sequence pn where pn
is the population of the world as
of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place
of the number e then an is a well-defined
sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined
recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
SEQUENCES
If we let an be the digit in the nth decimal place
of the number e then an is a well-defined
sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined
recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined
recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
FIBONACCI SEQUENCE
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
SEQUENCES
A sequence such as
that in Example 1 a
[an = n(n + 1)] can
be pictured either
by Plotting its terms on
a number line Plotting its graph
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
SEQUENCES
Note that since
a sequence is
a function whose
domain is the set
of positive integers
its graph consists of isolated points with
coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
Fig 1212 p 712
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
SEQUENCES
From either figure
it appears that
the terms of
the sequence
an = n(n + 1) are
approaching 1 as n
becomes large
Fig 1211 p 712
Fig 1212 p 712
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
SEQUENCES
In fact the difference
can be made as small as we like by taking n
sufficiently large
We indicate this by writing
11
1 1
n
n n
lim 11n
n
n
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
SEQUENCES
In general the notation
means that the terms of the sequence an
approach L as n becomes large
lim nna L
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
SEQUENCES
Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to L
as we like by taking n sufficiently large
If exists the sequence converges (or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
Definition 1
lim nna
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L
Fig 1213 p 713
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMIT OF A SEQUENCE
A more precise version of
Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding
integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure
in which the terms a1 a2 a3 are plotted
on a number line
Fig 1214 p 713
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε)
is chosen there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval
Fig 1214 p 713
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMIT OF A SEQUENCE
Another illustration of Definition 2
is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
Fig 1215 p 713
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMIT OF A SEQUENCE
This picture must be valid no matter how
small ε is chosen
Usually however a smaller ε requires a larger N
Fig 1215 p 713
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7
in Section 26 you will see that the only
difference between and
is that n is required to be an integer
Thus we have the following theorem
lim nna L
lim ( )
xf x L
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
If and f(n) = an when n
is an integer then
lim ( )x
f x L
lim nna L
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
Fig 1216 p 714
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
In particular since we know that
when r gt 0 (Theorem 5
in Section 26) we have
lim(1 ) 0r
xx
Equation 4
1lim 0 if 0
rnr
n
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
If an becomes large as n becomes large
we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMIT OF A SEQUENCE
means that for every positive
number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
If then the sequence an
is divergent but in a special way
We say that an diverges to infin
lim nna
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
The Limit Laws given in Section 23
also hold for the limits of sequences
and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent
sequences and
c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn n
nn n
n nn
PPn n n
n n
a b a b
aab
b b
a a p a
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
The Squeeze Theorem
can also be adapted for
sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and
thenlim nnb L
lim limn nn na c L
Fig 1217 p 715
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
Another useful fact about limits of
sequences is given by the following
theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
Theorem 6
lim 0nn
a
lim 0nna
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
n
n
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim
1 11 1 lim1 lim
11
1 0
n
n n
n n
n
nn n
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
n
n
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule
directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule
to the related function f(x) = (ln x)x and
obtain
Example 5
ln 1lim lim 0
1x x
x x
x
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
Therefore by Theorem 3
we have
Example 5
lnlim 0n
n
n
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
Determine whether
the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
If we write out the terms of the sequence
we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
Fig 1218 p 715
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)nn
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7
( 1)lim
n
n n
( 1) 1lim lim 0
n
n nn n
( 1)lim 0
n
n n
Fig 1219 p 716
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
The following theorem says that if we
apply a continuous function to the terms
of a convergent sequence the result is also
convergent
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
If and the function f is
continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
lim ( ) ( )nn
f a f L
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
n
lim sin( ) sin lim( )
sin 0 0n n
n n
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
Discuss the convergence
of the sequence an = nnn
where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
Both the numerator and denominator
approach infinity as n rarr infin
However here we have no corresponding
function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling
for what happens to an as n gets large
Example 9
1 2 3
1 2 1 2 31
2 2 3 3 3a a a
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
Therefore
1 2 3n
na
n n n n
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
From these expressions and the graph here
it appears that the terms are decreasing and
perhaps approach 0
Example 9
Fig 12110 p 716
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
To confirm this observe from Equation 8
that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
na
n n n n
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 9
1
n
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
For what values of r is the sequence r n
convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
lim 0x
xa
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3
we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
r
lim1 1 and lim 0 0n n
n n
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
lim 0n
nr
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in
Example 6
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
The figure shows the graphs for various
values of r
Example 10
Fig 12111 p 717
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlier
Example 10
Fig 1218 p 715
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
The results of Example 10
are summarized for future use
as follows
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
The sequence r n is convergent
if ndash1 lt r le 1 and divergent for all other
values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
r
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
DECREASING SEQUENCES
The sequence is decreasing
because
and so an gt an+1 for all n ge 1
Example 11
3
5n
3 3 3
5 ( 1) 5 6n n n
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1n
na
n
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
DECREASING SEQUENCES
We must show that an+1 lt an
that is
2 2
1
( 1) 1 1
n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication
2 22 2
3 2 3 2
2
1( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n nn n n n
n n
n n n n n n
n n
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
DECREASING SEQUENCES
Since n ge 1 we know that the inequality
n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
DECREASING SEQUENCES
Consider the function2
( ) 1
xf x
x
2 2
2 2
22
2 2
1 2( )
( 1)
10 whenever 1
( 1)
x xf x
x
xx
x
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le M for all n ge 1
Below if there is a number m such that m le an for all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
BOUNDED SEQUENCES
We know that not every bounded
sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
BOUNDED SEQUENCES
Similarly not every
monotonic sequence is
convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
BOUNDED SEQUENCES
However if a sequence is both bounded
and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
BOUNDED SEQUENCES
If an is increasing and an le M for all n
then the terms are forced to crowd together
and approach some number L
Fig 12112 p 718
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
BOUNDED SEQUENCES
The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
BOUNDED SEQUENCES
The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
Every bounded monotonic
sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
MONOTONIC SEQ THEOREM
Suppose an is an increasing
sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
Given ε gt 0 L ndash ε is not an upper bound
for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
MONOTONIC SEQ THEOREM
Thus
|L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
MONOTONIC SEQ THEOREM
A similar proof (using
the greatest lower bound)
works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
MONOTONIC SEQ THEOREM
This fact is used many times in
dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by
the recurrence relation
Example 13
11 1 22 ( 6) for 123n na a a n
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
MONOTONIC SEQ THEOREM
We begin by computing the first several
terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5
(5 6) 55 575 5875
59375 596875 5984375
a a a
a a a
a a a
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing
we use mathematical induction to show
that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because
a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = k
we have
Hence
and
Thus
Example 13
1k ka a
1 6 6k ka a 1 1
12 2( 6) ( 6)k ka a
2 1k ka a
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true
for n = k + 1
Therefore the inequality is true for all n
by induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by
showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka
6 12ka
Example 13
1 12 2( 6) (12) 6ka
1 6ka
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
MONOTONIC SEQ THEOREM
This shows by mathematical
induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing
and bounded Theorem 12 guarantees
that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know
exists we can use the recurrence relation
to write
lim nn
L a
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too
(as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L
Example 13