Post on 21-Jan-2018
The Harmonic Series and the Integral Test
The Harmonic Series and the Integral Test
If we add infinitely many terms and obtain a finite sum,
it must be the case that the terms get smaller and
smaller and goes to zero.
Theorem:
The Harmonic Series and the Integral Test
If we add infinitely many terms and obtain a finite sum,
it must be the case that the terms get smaller and
smaller and goes to zero.
If = a1 + a2 + a3 + … = L is a Σi=1
∞
ai
convergent series, then lim an = 0.n∞
Proof: Let = a1 + a2 .. = L be a convergent series. Σi=1
∞
ai
Theorem:
The Harmonic Series and the Integral Test
If we add infinitely many terms and obtain a finite sum,
it must be the case that the terms get smaller and
smaller and goes to zero.
If = a1 + a2 + a3 + … = L is a Σi=1
∞
ai
convergent series, then lim an = 0.n∞
Proof: Let = a1 + a2 .. = L be a convergent series. Σi=1
∞
ai
Theorem:
This means the for the sequence of partial sums,
lim sn = lim (a1 + a2 + … + an) = L converges.
The Harmonic Series and the Integral Test
If we add infinitely many terms and obtain a finite sum,
it must be the case that the terms get smaller and
smaller and goes to zero.
n∞
If = a1 + a2 + a3 + … = L is a Σi=1
∞
ai
convergent series, then lim an = 0.n∞
Proof: Let = a1 + a2 .. = L be a convergent series. Σi=1
∞
ai
Theorem:
The Harmonic Series and the Integral Test
If we add infinitely many terms and obtain a finite sum,
it must be the case that the terms get smaller and
smaller and goes to zero.
n∞
lim sn-1 = lim (a1 + a2 +..+ an-1) = L. n∞
If = a1 + a2 + a3 + … = L is a Σi=1
∞
ai
convergent series, then lim an = 0.n∞
On the other hand,
This means the for the sequence of partial sums,
lim sn = lim (a1 + a2 + … + an) = L converges.
Proof: Let = a1 + a2 .. = L be a convergent series. Σi=1
∞
ai
Theorem:
The Harmonic Series and the Integral Test
If we add infinitely many terms and obtain a finite sum,
it must be the case that the terms get smaller and
smaller and goes to zero.
n∞
lim sn-1 = lim (a1 + a2 +..+ an-1) = L.
Since an = sn – sn-1,n∞
If = a1 + a2 + a3 + … = L is a Σi=1
∞
ai
convergent series, then lim an = 0.n∞
On the other hand,
This means the for the sequence of partial sums,
lim sn = lim (a1 + a2 + … + an) = L converges.
Proof: Let = a1 + a2 .. = L be a convergent series. Σi=1
∞
ai
Theorem:
The Harmonic Series and the Integral Test
If we add infinitely many terms and obtain a finite sum,
it must be the case that the terms get smaller and
smaller and goes to zero.
n∞
lim sn-1 = lim (a1 + a2 +..+ an-1) = L.
Since an = sn – sn-1, so lim an = lim sn – sn-1 = L – L = 0.n∞
n∞
If = a1 + a2 + a3 + … = L is a Σi=1
∞
ai
convergent series, then lim an = 0.n∞
On the other hand,
This means the for the sequence of partial sums,
lim sn = lim (a1 + a2 + … + an) = L converges.
The Harmonic Series and the Integral Test
However the fact that lim an 0 does not guarantee
that their sum CGs to a finite number.
Example:
The sequence 1,
The Harmonic Series and the Integral Test
However the fact that lim an 0 does not guarantee
that their sum CGs to a finite number.
12 ,
12 ,
13 ,
13 ,
13 ,
14 ,
14 ,
14 ,
14 , 0, 1
5 , ..
Example:
The sequence 1,
The Harmonic Series and the Integral Test
However the fact that lim an 0 does not guarantee
that their sum CGs to a finite number.
12 ,
12 ,
13 ,
13 ,
13 ,
14 ,
14 ,
14 ,
14 , 0, 1
5 , ..
but their sum
Example:
The sequence 1,
The Harmonic Series and the Integral Test
However the fact that lim an 0 does not guarantee
that their sum CGs to a finite number.
12 ,
12 ,
13 ,
13 ,
13 ,
14 ,
14 ,
14 ,
14 , 0, 1
5 , ..
but their sum 1
Example:
The sequence 1,
The Harmonic Series and the Integral Test
However the fact that lim an 0 does not guarantee
that their sum CGs to a finite number.
12 ,
12 ,
13 ,
13 ,
13 ,
14 ,
14 ,
14 ,
14 , 0, 1
5 , ..
but their sum 1+ 12
+ 12
Example:
The sequence 1,
The Harmonic Series and the Integral Test
However the fact that lim an 0 does not guarantee
that their sum CGs to a finite number.
12 ,
12 ,
13 ,
13 ,
13 ,
14 ,
14 ,
14 ,
14 , 0, 1
5 , ..
but their sum 1+ 12
+ 12
13
13
13
+ + +
Example:
The sequence 1,
The Harmonic Series and the Integral Test
However the fact that lim an 0 does not guarantee
that their sum CGs to a finite number.
12 ,
12 ,
13 ,
13 ,
13 ,
14 ,
14 ,
14 ,
14 , 0, 1
5 , ..
but their sum 1+ 12
+ 12
13
13
13
14
14
14
14
+ + + + + + +
Example:
The sequence 1,
The Harmonic Series and the Integral Test
However the fact that lim an 0 does not guarantee
that their sum CGs to a finite number.
12 ,
12 ,
13 ,
13 ,
13 ,
14 ,
14 ,
14 ,
14 , 0, 1
5 , ..
but their sum 1+ 12
+ 12
13
13
13
14
14
14
14
15 ..+ + + + + + + + = ∞
Example:
The sequence 1,
The Harmonic Series and the Integral Test
However the fact that lim an 0 does not guarantee
that their sum CGs to a finite number.
12 ,
12 ,
13 ,
13 ,
13 ,
14 ,
14 ,
14 ,
14 , 0, 1
5 , ..
but their sum 1+ 12
+ 12
13
13
13
14
14
14
14
15 ..+ + + + + + + + = ∞
An important sequence that goes to 0 but sums to ∞
is the harmonic sequence: {1/n} = 12 ,
13 ,
14 , ..1,{ }
Example:
The sequence 1,
The Harmonic Series and the Integral Test
However the fact that lim an 0 does not guarantee
that their sum CGs to a finite number.
12 ,
12 ,
13 ,
13 ,
13 ,
14 ,
14 ,
14 ,
14 , 0, 1
5 , ..
but their sum 1+ 12
+ 12
13
13
13
14
14
14
14
15 ..+ + + + + + + + = ∞
An important sequence that goes to 0 but sums to ∞
is the harmonic sequence: {1/n} = 12 ,
13 ,
14 , ..1,{ }
To see that they sum to ∞, sum in blocks as shown:
Example:
The sequence 1,
The Harmonic Series and the Integral Test
However the fact that lim an 0 does not guarantee
that their sum CGs to a finite number.
12 ,
12 ,
13 ,
13 ,
13 ,
14 ,
14 ,
14 ,
14 , 0, 1
5 , ..
but their sum 1+ 12
+ 12
13
13
13
14
14
14
14
15 ..+ + + + + + + + = ∞
An important sequence that goes to 0 but sums to ∞
is the harmonic sequence: {1/n} = 12 ,
13 ,
14 , ..1,{ }
To see that they sum to ∞, sum in blocks as shown:
1
Example:
The sequence 1,
The Harmonic Series and the Integral Test
However the fact that lim an 0 does not guarantee
that their sum CGs to a finite number.
12 ,
12 ,
13 ,
13 ,
13 ,
14 ,
14 ,
14 ,
14 , 0, 1
5 , ..
but their sum 1+ 12
+ 12
13
13
13
14
14
14
14
15 ..+ + + + + + + + = ∞
An important sequence that goes to 0 but sums to ∞
is the harmonic sequence: {1/n} = 12 ,
13 ,
14 , ..1,{ }
To see that they sum to ∞, sum in blocks as shown:
12
13
110
...1 + + + +
Example:
The sequence 1,
The Harmonic Series and the Integral Test
However the fact that lim an 0 does not guarantee
that their sum CGs to a finite number.
12 ,
12 ,
13 ,
13 ,
13 ,
14 ,
14 ,
14 ,
14 , 0, 1
5 , ..
but their sum 1+ 12
+ 12
13
13
13
14
14
14
14
15 ..+ + + + + + + + = ∞
An important sequence that goes to 0 but sums to ∞
is the harmonic sequence: {1/n} = 12 ,
13 ,
14 , ..1,{ }
To see that they sum to ∞, sum in blocks as shown:
12
13
110
...1 + + + +
> 910
Example:
The sequence 1,
The Harmonic Series and the Integral Test
However the fact that lim an 0 does not guarantee
that their sum CGs to a finite number.
12 ,
12 ,
13 ,
13 ,
13 ,
14 ,
14 ,
14 ,
14 , 0, 1
5 , ..
but their sum 1+ 12
+ 12
13
13
13
14
14
14
14
15 ..+ + + + + + + + = ∞
An important sequence that goes to 0 but sums to ∞
is the harmonic sequence: {1/n} = 12 ,
13 ,
14 , ..1,{ }
To see that they sum to ∞, sum in blocks as shown:
12
13
110
...1 + + + + 111
+ 1100
... +
> 910
Example:
The sequence 1,
The Harmonic Series and the Integral Test
However the fact that lim an 0 does not guarantee
that their sum CGs to a finite number.
12 ,
12 ,
13 ,
13 ,
13 ,
14 ,
14 ,
14 ,
14 , 0, 1
5 , ..
but their sum 1+ 12
+ 12
13
13
13
14
14
14
14
15 ..+ + + + + + + + = ∞
An important sequence that goes to 0 but sums to ∞
is the harmonic sequence: {1/n} = 12 ,
13 ,
14 , ..1,{ }
To see that they sum to ∞, sum in blocks as shown:
12
13
110
...1 + + + + 111
+ 1100
... +
> 910 > 90
100
Example:
The sequence 1,
The Harmonic Series and the Integral Test
However the fact that lim an 0 does not guarantee
that their sum CGs to a finite number.
12 ,
12 ,
13 ,
13 ,
13 ,
14 ,
14 ,
14 ,
14 , 0, 1
5 , ..
but their sum 1+ 12
+ 12
13
13
13
14
14
14
14
15 ..+ + + + + + + + = ∞
An important sequence that goes to 0 but sums to ∞
is the harmonic sequence: {1/n} = 12 ,
13 ,
14 , ..1,{ }
To see that they sum to ∞, sum in blocks as shown:
12
13
110
...1 + + + + 111
+ 1100
... +
> 910 > 90
100= 9
10
Example:
The sequence 1,
The Harmonic Series and the Integral Test
However the fact that lim an 0 does not guarantee
that their sum CGs to a finite number.
12 ,
12 ,
13 ,
13 ,
13 ,
14 ,
14 ,
14 ,
14 , 0, 1
5 , ..
but their sum 1+ 12
+ 12
13
13
13
14
14
14
14
15 ..+ + + + + + + + = ∞
An important sequence that goes to 0 but sums to ∞
is the harmonic sequence: {1/n} = 12 ,
13 ,
14 , ..1,{ }
To see that they sum to ∞, sum in blocks as shown:
12
13
110
...1 + + + + 111
+ 1100
... + 1101
+ 11000
... +
> 910 > 90
100= 9
10
Example:
The sequence 1,
The Harmonic Series and the Integral Test
However the fact that lim an 0 does not guarantee
that their sum CGs to a finite number.
12 ,
12 ,
13 ,
13 ,
13 ,
14 ,
14 ,
14 ,
14 , 0, 1
5 , ..
but their sum 1+ 12
+ 12
13
13
13
14
14
14
14
15 ..+ + + + + + + + = ∞
An important sequence that goes to 0 but sums to ∞
is the harmonic sequence: {1/n} = 12 ,
13 ,
14 , ..1,{ }
To see that they sum to ∞, sum in blocks as shown:
12
13
110
...1 + + + + 111
+ 1100
... + 1101
+ 11000
... +
> 910 > 90
100= 9
10>
1000900
Example:
The sequence 1,
The Harmonic Series and the Integral Test
However the fact that lim an 0 does not guarantee
that their sum CGs to a finite number.
12 ,
12 ,
13 ,
13 ,
13 ,
14 ,
14 ,
14 ,
14 , 0, 1
5 , ..
but their sum 1+ 12
+ 12
13
13
13
14
14
14
14
15 ..+ + + + + + + + = ∞
An important sequence that goes to 0 but sums to ∞
is the harmonic sequence: {1/n} = 12 ,
13 ,
14 , ..1,{ }
To see that they sum to ∞, sum in blocks as shown:
12
13
110
...1 + + + + 111
+ 1100
... + 1101
+ 11000
... +
> 910 > 90
100= 9
10>
1000=
10900 9
Example:
The sequence 1,
The Harmonic Series and the Integral Test
However the fact that lim an 0 does not guarantee
that their sum CGs to a finite number.
12 ,
12 ,
13 ,
13 ,
13 ,
14 ,
14 ,
14 ,
14 , 0, 1
5 , ..
but their sum 1+ 12
+ 12
13
13
13
14
14
14
14
15 ..+ + + + + + + + = ∞
An important sequence that goes to 0 but sums to ∞
is the harmonic sequence: {1/n} = 12 ,
13 ,
14 , ..1,{ }
To see that they sum to ∞, sum in blocks as shown:
12
13
110
...1 + + + + 111
+ 1100
... + 1101
+ 11000
... + + …
> 910 > 90
100= 9
10>
1000=
10900 9
= ∞
Example:
The sequence 1,
The Harmonic Series and the Integral Test
However the fact that lim an 0 does not guarantee
that their sum CGs to a finite number.
12 ,
12 ,
13 ,
13 ,
13 ,
14 ,
14 ,
14 ,
14 , 0, 1
5 , ..
but their sum 1+ 12
+ 12
13
13
13
14
14
14
14
15 ..+ + + + + + + + = ∞
An important sequence that goes to 0 but sums to ∞
is the harmonic sequence: {1/n} = 12 ,
13 ,
14 , ..1,{ }
To see that they sum to ∞, sum in blocks as shown:
12
13
110
...1 + + + + 111
+ 1100
... + 1101
+ 11000
... + + …
> 910 > 90
100= 9
10>
1000=
10900 9
= ∞
Hence the harmonic series DGs.
The Harmonic Series and the Integral Test
The following theorem and theorems in the next
section give various methods of determining if a
series is convergent or divergent.
The Harmonic Series and the Integral Test
The following theorem and theorems in the next
section give various methods of determining if a
series is convergent or divergent.
We shall assume all series are positive series, i.e.
all terms in the series are positive unless stated
otherwise.
Σi=1
∞
ai
Theorem:
The Harmonic Series and the Integral Test
The following theorem and theorems in the next
section give various methods of determining if a
series is convergent or divergent.
(Integral Test) If an = f(n) > 0, then
CGs if and only if
We shall assume all series are positive series, i.e.
all terms in the series are positive unless stated
otherwise.
∫1 f(x) dx CGs. ∞
Σi=1
∞
ai
Theorem:
The Harmonic Series and the Integral Test
The following theorem and theorems in the next
section give various methods of determining if a
series is convergent or divergent.
(Integral Test) If an = f(n) > 0, then
CGs if and only if
We shall assume all series are positive series, i.e.
all terms in the series are positive unless stated
otherwise.
∫1 f(x) dx CGs. ∞
Combine this with the p-theorem from before, we
have the following theorem about the convergence of
the p-series:
Σi=1
Theorem:
The Harmonic Series and the Integral Test
(p-series) CGs if and only if p > 1. ∞
np1
Σi=1
Theorem:
The Harmonic Series and the Integral Test
(p-series) CGs if and only if p > 1. ∞
np1
Proof:
Σi=1
CGs if and only if CGs. ∞
np1
By the integral test,
∫1 x
p1
∞
dx
Σi=1
Theorem:
The Harmonic Series and the Integral Test
(p-series) CGs if and only if p > 1. ∞
np1
Proof: By the integral test,
By the p-theorem, this integral CGs if and only if p >1.
Σi=1
CGs if and only if CGs. ∞
np1
∫1 x
p1
∞
dx
Σi=1
Theorem:
The Harmonic Series and the Integral Test
(p-series) CGs if and only if p > 1. ∞
np1
Proof: By the integral test,
By the p-theorem, this integral CGs if and only if p >1.
So CGs if and only if p > 1. Σi=1
∞
np1
Σi=1
CGs if and only if CGs. ∞
np1
∫1 x
p1
∞
dx
Σi=1
Theorem:
The Harmonic Series and the Integral Test
(p-series) CGs if and only if p > 1. ∞
np1
Proof: By the integral test,
By the p-theorem, this integral CGs if and only if p >1.
So CGs if and only if p > 1.
Example:
a. Σi=1
∞
n3/21
b. Σi=1
∞
n1
Σi=1
∞
np1
Σi=1
CGs if and only if CGs. ∞
np1
∫1 x
p1
∞
dx
Σi=1
Theorem:
The Harmonic Series and the Integral Test
(p-series) CGs if and only if p > 1. ∞
np1
Proof: By the integral test,
By the p-theorem, this integral CGs if and only if p >1.
So CGs if and only if p > 1.
Example:
a. CGs since 3/2 > 1.Σi=1
∞
n3/21
b. Σi=1
∞
n1
Σi=1
∞
np1
Σi=1
CGs if and only if CGs. ∞
np1
∫1 x
p1
∞
dx
Σi=1
Theorem:
The Harmonic Series and the Integral Test
(p-series) CGs if and only if p > 1. ∞
np1
Proof: By the integral test,
By the p-theorem, this integral CGs if and only if p >1.
So CGs if and only if p > 1.
Example:
a. CGs since 3/2 > 1.Σi=1
∞
n3/21
b. DGs since 1/2 < 1.Σi=1
∞
n1
Σi=1
∞
np1
Σi=1
CGs if and only if CGs. ∞
np1
∫1 x
p1
∞
dx
Σi=1
Theorem:
The Harmonic Series and the Integral Test
(p-series) CGs if and only if p > 1. ∞
np1
Proof: By the integral test,
By the p-theorem, this integral CGs if and only if p >1.
So CGs if and only if p > 1.
Example:
a. CGs since 3/2 > 1.Σi=1
∞
n3/21
b. DGs since 1/2 < 1.Σi=1
∞
n1
This theorem applies to series that are p-series
except for finitely many terms (eventual p-series).
Σi=1
∞
np1
Σi=1
CGs if and only if CGs. ∞
np1
∫1 x
p1
∞
dx
Recall the following theorems of improper integrals.
(The Floor Theorem)
The Harmonic Series and the Integral Test
(The Floor Theorem)
y = f(x)
y = g(x)∞
The Harmonic Series and the Integral Test
(The Floor Theorem)
If f(x) > g(x) > 0 and g(x) dx = ∞, ∫a
b
y = f(x)
y = g(x)∞
The Harmonic Series and the Integral Test
(The Floor Theorem)
If f(x) > g(x) > 0 and g(x) dx = ∞, then f(x) = ∞. ∫a
b
∫a
b
y = f(x)
y = g(x)∞
The Harmonic Series and the Integral Test
(The Floor Theorem)
If f(x) > g(x) > 0 and g(x) dx = ∞, then f(x) = ∞. ∫a
b
∫a
b
y = f(x)
y = g(x)∞
(The Ceiling theorem)
The Harmonic Series and the Integral Test
(The Floor Theorem)
If f(x) > g(x) > 0 and g(x) dx = ∞, then f(x) = ∞. ∫a
b
∫a
b
y = f(x)
y = g(x)∞
(The Ceiling theorem)
y = f(x)
y = g(x)
N
The Harmonic Series and the Integral Test
(The Floor Theorem)
If f(x) > g(x) > 0 and g(x) dx = ∞, then f(x) = ∞. ∫a
b
∫a
b
y = f(x)
y = g(x)∞
(The Ceiling theorem)
If f(x) > g(x) > 0 and f(x) dx = N converges∫a
b
y = f(x)
y = g(x)
N
The Harmonic Series and the Integral Test
(The Floor Theorem)
If f(x) > g(x) > 0 and g(x) dx = ∞, then f(x) = ∞. ∫a
b
∫a
b
y = f(x)
y = g(x)∞
(The Ceiling theorem)
If f(x) > g(x) > 0 and f(x) dx = N converges then
g(x) dx converges also.
∫a
b
∫a
b
y = f(x)
y = g(x)
N
The Harmonic Series and the Integral Test
The Harmonic Series and the Integral Test
By the same logic we have their discrete versions.
The Harmonic Series and the Integral Test
By the same logic we have their discrete versions.
The Floor Theorem
The Harmonic Series and the Integral Test
By the same logic we have their discrete versions.
The Floor Theorem
Let {an} and {bn} be two sequences and an > bn > 0.
The Harmonic Series and the Integral Test
Suppose bn = ∞, then an = ∞. Σi=k
∞
Σi=k
∞
By the same logic we have their discrete versions.
The Floor Theorem
Let {an} and {bn} be two sequences and an > bn > 0.
The Harmonic Series and the Integral Test
Suppose bn = ∞, then an = ∞. Σi=k
∞
Σi=k
∞
Example: Does CG or DG?
By the same logic we have their discrete versions.
Σi=2
∞
Ln(n) 1
The Floor Theorem
Let {an} and {bn} be two sequences and an > bn > 0.
The Harmonic Series and the Integral Test
Suppose bn = ∞, then an = ∞. Σi=k
∞
Σi=k
∞
Example: Does CG or DG?
For n > 1, n > Ln(n), (why?)
By the same logic we have their discrete versions.
Σi=2
∞
Ln(n) 1
The Floor Theorem
Let {an} and {bn} be two sequences and an > bn > 0.
The Harmonic Series and the Integral Test
Suppose bn = ∞, then an = ∞. Σi=k
∞
Σi=k
∞
Example: Does CG or DG?
Ln(n) 1 >
n . 1
For n > 1, n > Ln(n), (why?)
By the same logic we have their discrete versions.
Σi=2
∞
Ln(n) 1
so
The Floor Theorem
Let {an} and {bn} be two sequences and an > bn > 0.
The Harmonic Series and the Integral Test
Suppose bn = ∞, then an = ∞. Σi=k
∞
Σi=k
∞
Example: Does CG or DG?
Ln(n) 1 >
n . 1
For n > 1, n > Ln(n), (why?)
Σi=2 n
1Hence Σi=2 Ln(n)
2
By the same logic we have their discrete versions.
>
Σi=2
∞
Ln(n) 1
so
The Floor Theorem
Let {an} and {bn} be two sequences and an > bn > 0.
The Harmonic Series and the Integral Test
Suppose bn = ∞, then an = ∞. Σi=k
∞
Σi=k
∞
Example: Does CG or DG?
Ln(n) 1 >
n . 1
For n > 1, n > Ln(n), (why?)
Σi=2 n
1Hence Σi=2 Ln(n)
2
By the same logic we have their discrete versions.
> = ∞ because it’s harmonic.
Σi=2
∞
Ln(n) 1
so
The Floor Theorem
Let {an} and {bn} be two sequences and an > bn > 0.
The Harmonic Series and the Integral Test
Suppose bn = ∞, then an = ∞. Σi=k
∞
Σi=k
∞
Example: Does CG or DG?
Ln(n) 1 >
n . 1
For n > 1, n > Ln(n), (why?)
Σi=2 n
1
Therefore
Hence Σi=2 Ln(n)
2
By the same logic we have their discrete versions.
> = ∞ because it’s harmonic.
Σi=2
∞
Ln(n) 1
Σi=2 Ln(n)
2
so
= ∞ or that it DGs.
The Floor Theorem
Let {an} and {bn} be two sequences and an > bn > 0.
The Harmonic Series and the Integral Test
Suppose bn = ∞, then an = ∞. Σi=k
∞
Σi=k
∞
Example: Does CG or DG?
Ln(n) 1 >
n . 1
For n > 1, n > Ln(n), (why?)
Σi=2 n
1
Therefore
Hence Σi=2 Ln(n)
2
By the same logic we have their discrete versions.
> = ∞ because it’s harmonic.
Σi=2
∞
Ln(n) 1
Σi=2 Ln(n)
2
so
= ∞ or that it DGs.
The Floor Theorem
Let {an} and {bn} be two sequences and an > bn > 0.
Note that no conclusion can be drawn about Σan if that
Σ bn < ∞ i.e. Σ an may CG or it may DG. (Why so?)
The Harmonic Series and the Integral Test
The Ceiling Theorem
The Harmonic Series and the Integral Test
The Ceiling Theorem
Let {an} and {bn} be two sequences and an > bn > 0.
The Harmonic Series and the Integral Test
Suppose that an CGs, then bn CGs.Σi=k
Σi=k
The Ceiling Theorem
Let {an} and {bn} be two sequences and an > bn > 0.
The Harmonic Series and the Integral Test
Suppose that an CGs, then bn CGs.Σi=k
Σi=k
Example: Does CG or DG?
The Ceiling Theorem
Let {an} and {bn} be two sequences and an > bn > 0.
Σi=1
∞
n2 + 4 2
The Harmonic Series and the Integral Test
Suppose that an CGs, then bn CGs.Σi=k
Σi=k
Example: Does CG or DG?
The Ceiling Theorem
Let {an} and {bn} be two sequences and an > bn > 0.
Σi=1
∞
n2 + 4 2
Compare with n2 + 4
2n22
The Harmonic Series and the Integral Test
Suppose that an CGs, then bn CGs.Σi=k
Σi=k
Example: Does CG or DG?
n2 + 4 2>
n22
. we have
The Ceiling Theorem
Let {an} and {bn} be two sequences and an > bn > 0.
Σi=1
∞
n2 + 4 2
Compare with n2 + 4
2n22
The Harmonic Series and the Integral Test
Suppose that an CGs, then bn CGs.Σi=k
Σi=k
Example: Does CG or DG?
n2 + 4 2>
n22
.
Σ n22
we have
The Ceiling Theorem
Let {an} and {bn} be two sequences and an > bn > 0.
Σi=1
∞
n2 + 4 2
Compare with n2 + 4
2n22
= 2Σ n21
The Harmonic Series and the Integral Test
Suppose that an CGs, then bn CGs.Σi=k
Σi=k
Example: Does CG or DG?
n2 + 4 2>
n22
.
Σ n22
we have
CGs since it’s the p–series with p = 2 > 1,
The Ceiling Theorem
Let {an} and {bn} be two sequences and an > bn > 0.
Σi=1
∞
n2 + 4 2
Compare with n2 + 4
2n22
= 2Σ n21
The Harmonic Series and the Integral Test
Suppose that an CGs, then bn CGs.Σi=k
Σi=k
Example: Does CG or DG?
n2 + 4 2>
n22
.
Σ n22
we have
CGs since it’s the p–series with p = 2 > 1,
n2 + 4
The Ceiling Theorem
Let {an} and {bn} be two sequences and an > bn > 0.
Σi=1
∞
n2 + 4 2
Compare with n2 + 4
2n22
2we see that Σ CGs also.
= 2Σ n21
The Harmonic Series and the Integral Test
Suppose that an CGs, then bn CGs.Σi=k
Σi=k
Example: Does CG or DG?
n2 + 4 2>
n22
.
Σ n22
we have
CGs since it’s the p–series with p = 2 > 1,
n2 + 4
The Ceiling Theorem
Let {an} and {bn} be two sequences and an > bn > 0.
Σi=1
∞
n2 + 4 2
Compare with n2 + 4
2n22
2
Note that no conclusion can be drawn about Σbn if that
Σan = ∞ i.e. Σ bn may CG or it may DG. (Why so?)
we see that Σ CGs also.
= 2Σ n21