Topic4 Series

8/6/2019 Topic4 Series

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Series

1.0 : Introduction

Here we cover the basics of series. This will include :

251.8.5 : Test 3 : Ratio test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

231.8.4 : Proof of the comparison test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

211.8.3 : Test 2 : Comparison test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

201.8.2 : Proof of limit test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

201.8.1 : Test 1 : Limit test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

191.8 : Test for Convergence of Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

161.7 : Method of Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

151.6 : Summation of Number Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .131.5.3 : The Arithmetic-Geometric mean inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

121.5.2 : Geometric means . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

111.5.1 : Arithmetic means . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

111.5 : Arithmetic and geometric means (private reading/self study) . . . . . . . . . . . . . . . . . . . . .

101.4 : Convergence of a Series and the Sum to Infinity of a Geometric Progressions . . . . . . .

81.3.2 : The sum of a geometric progression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

71.3.1 : Basic aspects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

71.3 : Geometric Progressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

41.2.2 : The sum of an arithmetic progression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

31.2.1 : Basic aspects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

31.2 : Arithmetic Progressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .21.1.3 : Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

21.1.2 : Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

21.1.1 : Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

21.1 : Basic Issues : Sequences, Series and Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11.0 : Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

www.ucl.ac.uk/~uczlcfe

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1.1 : Basic Issues : Sequences, Series and Notation

Here we will summarise what sequences and series are, and the notation used in the study of these.

1.1.1 : Sequences

Consider, therefore, the following set of numbers :

1) 2, 3, 4, 5, ....... 2) 2, 4, 6, 8, 10, ...... 3) 4, 9, 16, 25, ......

Informally we may say that : in 1) each number is one more than the previous number

in 2) each number is twice the previous number

in 3) each number is the square of the natural numbers 1, 2, 3, ....

Formally we may then define a sequence as :

a set of terms (numbers, variables, etc...) in a specific

order with a rule for obtaining these terms

1.1.2 : Series

Consider now the following type of sequence

4) 2 + 3 + 4 + 5 + .......

5) 2 + 4 + 6 + 8 + 10 ......

6) 4 + 9 + 16 + 25 + ......

Here we have added up all the terms of the sequences 1), 2), and 3). Formally we may then generally

define a series as :

the sum of terms of a sequence

The specific formal description for each of the series above can then be stated as :

for 4) forr+ 1 r= 1,2,3,...for 5) for2r r= 1,2,3,...for 6) forr2 r= 2,3,4,...

1.1.3 : Notation

The above formalisations of the series may be improved by using the sigma notation . Hence,

for 4) for 5) for 6)r=1

r+ 1 r=1

2r r=2

r2

Of course we may define a series to have a finite, fixed, length, i.e. . in this seriesr=4

6

r2 = 42 + 52 + 62

the 1st term is 4, the 2nd term is 5, and the 3rd term is 6.


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1.2 : Arithmetic Progressions

Here we will study a specific type of series where the sameconstantnumber is added from one term to

the next. This implies that the difference between each term is constant.

1.2.1 : Basic aspectsConsider then the sequence,

5, 8, 11, 14, ....., 29

Informally we may describe this sequence as having a 1st term of 5, 2nd term of 8, 3rd term of 11, 4th

term of 14, and nth term of 29. We notice that the difference between each number is a constant 3, and

therefore does not change.

We may now re-rewrite the above sequence in terms of this constant difference, i.e.

5, (5 + 3), (5 + 2x3), (5 + 3x3), ....., (5 + 8x3)

Formalising the above description we may then say that

the 1st term is a

the 2nd term is a + d

the 3rd term is a + 2d

...................... ..........

the nth term is a + (n-1).d

where d is called the common difference. Such a sequence of terms is called an arithmetic

progression and it is arithmetic because of the constant common difference. In general the commondifference can be found easily by subtracting one term by the previous term :

d= (a + k.d) (a + (k 1).d)

Examples

See lecture.

Exercise on reading mathematically

By referring to any textbook, and/or the notes above, interpret fully the concept of arithmetic

progressions and their formalisation (i.e. how they are described mathematically). If necessary, use the

examples done in class in order to help you refer to the aspects of arithmetic progressions. How does

each term follow on from the previous term ? How does each term continue onto the next term ? How

is each term similar and/or different from any otherterm ?

End of Exercise


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1.2.2 : The sum of an arithmetic progression

Given any particular arithmetic series (i.e. an arithmetic sequence whose terms are added together) how

do we go about finding its sum ? In other words what is the result of actually evaluating the series ?

Consider the arithmetic series

2 + 4 + 6 + ..... + 18 + 20

It can be tedious to manually add these terms in order to find the result. Furthermore, if there are 100s

of terms then it becomes impractical to do this. We therefore need a formula for being able to easily

find the sum of these terms.

Consider therefore re-writing the above series backwards and adding it to the above series, i.e.

S = 2 + 4 + 6 + 8 +..... + 18 + 20

S = 20 + 18 + 16 + 14 + ..... + 4 + 2

__________________________________

2S = 22 + 22 + 22 + 22 + ...... + 22 + 22Hence

2S = 10x22

So S = 110

We can generalise this method to a general arithmetic series which has 1st term a, common difference

d, and last term l :

Sn = a + (a + d) + (a + 2d) + ...... + (l - d) + l

Sn = l + (l - d) + (l - 2d) + ......+ (a + d) + a

___________________________________________________________

2Sn = (a + l) + (a + l) + (a + l) + ...... + (a + l) + (a + l)

Hence

(1.2.1)2Sn = n.(a + l)

But we know also that the last term l is a+(n-1).d, so substituting this into (1.2.1), and doing some

algebra, we get

(1.2.2)Sn =n2

[2a + (n 1).d]

We can now state one standard result about series, namely the sum of the natural integers from 1 to n

(i.e. 1+2+3+4+5+....) :

(1.2.3)i=1

n

i =n2

(n + 1)

This can be derived from (1.2.2) knowing that a = 1 and d = 1. Similarly we can show that

(1.2.4)i=1

n

2i =n2

(2n + 2)


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where, in this series, a = 2 and d = 2. Similarly we can show that

(1.2.5)i=1

n

3i =n2

(3n + 3)

and in general we have

(1.2.6)i=1

n

ki =n2

(kn + k)

for any constant k. We can formally proves this by :

saying that since kis constant we can factorise it out of the sum and therefore we can write

i=1

n

ki = k.i=1

n

i

Since we know from (1.2.3) that we can see that (6) is true.i=1n i = n/2(n + 1)

Examples

See lecture.

Exercises

See exercise sheets


1) Interpret the whole proof which leads to the formula (2) above. Describe fully the development

of the proof above, i.e. describe each step, and/or part of a step, in the proof above.

2) Interpret equation (2) above. Describe fully in plain English, and in your own words, the

expression as a whole as well as the separate parts of the expression. Also describe how you went about

interpreting the expression : this does not mean that you should repeat your interpretation, but that you

should describe what you did in order to get the interpretation you got. How did you read the

expression (2) in order to interpret it the way you did ? Was there any specific ways of reading that you

used in order to describe (2) as you did ? If so, what specific ways did you read (2) in ?

3) In equation 2) on the previous page, what does d=0 mean ? What is the effect on the serieswhen d=0 ? What kind of series would we have if d=0 ?

You may either interpret from the notes above, from any textbook, or from any other material you

wish. If you interpret from other material be sure to make a copy for yourselves for revision purposes.

End of Exercise


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Exercise on mathematical thinking

Q : What is it about the idea of writing the series backwards that allows us to derive the formula (2) ?

What is it about the way in which the terms of the backward and forward series are added together

which allows for the formula to be developed ?

Separately, note that in developing the formula (2) above the whole backward series is aligned

directly under the forward series. Supposing instead that when we write the series backwards we shift

its position by 1 place. Will we get a formula which gives the same result as (2) ?. So, supposing we do

S = 2 + 4 + 6 + 8

S = 8 + 6 + 4 + 2

____________________

2S = 2 + 12+12+12+ 2

Hence we get2S = 2x2 + 3x12

and formally

2S = 2.(1st term + last term) + 12.(n - 2)

Since the last term is the same as the 1st term we can say

===> (i)2S = 2(a + a) + 12(n 2) S = 2a + 6(n 2)

Q : Does (i) give the same result as (2) for the series 2 + 4 + 6 + 8 ?

By formalising the process above, develop an expression for a general arithmetic series containing 5

terms of 1st term a and common difference d, i.e. from

Sn = a + (a + d) + (a + 2d) + (a + 3d)

Sn = (a + 3d) + (a + 2d) + (a + d) + a

develop a general formula for this arithmetic progression.

Repeat the numerical example above when the backward series is shifted by 2 places, 3 places, etc...,

and again use mathematical analysis to develop general formulas for these cases.

Q : Do your formulae give the same result as (2) for the series 2 + 4 + 6 + 8 ? Do they give the same

result as (2) for any series ? Confirm that this is true or not by using any series of your choice.

End of Exercise


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1.3.2 : The sum of a geometric progression

Given any particular geometric series (i.e. a geometric sequence whose terms are added together) how

do we go about finding its sum ? In other words what is the result of actually evaluating the series ?

Consider the geometric series

1 + 3 + 32

+ 33

..... + 36

+ 37

It can be tedious to manually add these terms in order to find the result. Furthermore, if there are 100s

of terms then it becomes impractical to do this. We therefore need a formula for being able to easily

find the sum of these terms.

Consider therefore multiplying the series by 3 (the common ratio) and subtracting them :

S = 1 + 3 + 3 + 33 + 34 + ..... + 36 + 37

3S = 3 + 3 + 33 + 34 + ..... + 36 + 37 + 38

____________________________________________________

S - 3S = 1 + 0 + 0 + 0 + 0 + ..... + 0 + 0 - 38

(Note how the 2nd series is positioned : it is shifted by 1 place). Hence

===>S(1 3) = 1 38 S =1 38

(1 3)

We can generalise this method to a general geometric series which has 1st term a, and common ratio r.

Consider therefore multiplying the series by r(the common ratio) and subtracting them :

Sn = a + a.r + a.r + a.r

3

+ ..... + a.r

n-1

r.Sn = a.r + a.r + a.r3 + ..... + a.rn-1 + a.rn

___________________________________________________________

S - r.S = a + 0 + 0 + 0 + ..... + 0 - a.r n

Hence

===> (7)Sn(1 r) = a(1 rn) Sn =a(1 rn )

(1 r)

Expression (7) is then the standard formula for the sum of a geometric series.


1) Interpret the whole proof which leads to the formula (7) above. Describe fully the development

of the proof above, i.e. describe each step, and/or part of a step, in the proof above.

2) Interpret equation (7) above. Describe fully in plain English, and in your own words, the

expression as a whole as well as the separate parts of the expression. Also describe how you went about

interpreting the expression : this does not mean that you should repeat your interpretation, but that you

should describe what you did in order to get the interpretation you got.


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How did you read the expression (7) in order to interpret it the way you did ? Was there any specific

ways of reading that you used in order to describe (7) as you did ? If so, what specific ways did you

read (7) in ?

You may either interpret from the notes above, from any textbook, or from any other material you

wish. If you interpret from other material be sure to make a copy for yourselves for revision purposes.

End of Exercise


Q : Note that in developing the formula (7) above the series which is multiplied by r is shifted by 1

place before doing the subtraction. What is it about the idea of shifting the series forward by 1 place

that allows us to derive the formula (7) ? What is it about the way in which the terms of the shifted

series are subtracted which allows for the formula to be developed ?

Supposing instead that when we write the 2nd series we dont shift it at all. Will we get a formula

which gives the same result as (7) ?. So, supposing we do

S = 1 + 3 + 3 + 33 + 34 + ..... + 36 + 37

3S = 3 + 3 + 33 + 34 + 35 + ..... + 37 + 38

____________________________________________________

S - 3S = 1-3 + 3-3 + 3-33 + ........

Hence we get

-2S = 1-3 + 3.(1-3) +3(1-3) +33(1-3) + ....

-2S = (1-3).(1 + 3 + 3 + 33 + ....)

etc... What happens when we complete this analysis ?

Furthermore, by formalising the process above, develop an analysis of a general geometric series

containing 5 terms of 1st term a and common ratio r, i.e. from

Sn = a + a.r + a.r + a.r3 + ..... + a.rn-1

r.Sn = a.r + a.r + a.r3 + ..... + a.rn-1 + a.rn

analyse the effect of this structure on the development of any formula (if any).

Repeat the numerical example above when the series is shifted by 2 places, 3 places, etc..., and again

use mathematical analysis to develop general formulas for these cases.

Q : Do your formulae give the same result as (7) for the series 1 + 3 + 9 + 27? Do they give the same

result as (7) for any series ? Confirm that this is true or not by using any series of your choice.

End of Exercise


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Examples

See lecture.

Exercises

See exercise sheets

1.4 : Convergence of a Series and the Sum to Infinity of a Geometric Progressions

In the previous two sections we have studied the sum of series consisting of fixed number of terms.

What happens if we have an infinite number of term ? How does this affect the final answer we get

from an arithmetic or geometric series ? Will the series tend towards a finite answers, i.e. will it

converge ? Or will the series give us answer which is infinite, i.e. will it diverge ?

To understand the concept of convergence refer to the example discussed in the book on p599. In

summary, we may take a piece of string, of fixed length l, and cut it in half. We may continue cutting in

half any one of the parts we have left, and we can (theoretically) continue doing this forever. Thiscontinual cutting in half can be represented as a geometric series :

l2

+l4

+l8

+l

16+ ....

or, with r = l2

+ l22

+ l23

+ l24

+ .... + l2n

Although the series carries on forever, we know that it should add up to the total length of the string, l.

Informally we may say that the series converges to l, and formally we have that :

limnd l2 + l22 + l23 + l24 + .... + l2n = l

Not all series converge. Let us recap on the two formulae for the sum of arithmetic and geometric

progressions :

for an arithmetic progression (A.P.) of 1st term a, last term n, common difference dwe have

Sn =n2

[2a + (n 1).d]

for a geometric progression (G.P.) of 1st term a, and common ratio rwe have

Sn =a(1 rn )

(1 r)

Let us therefore consider the A.P. formula above. The question we need to answer is : what is the main

term which controls the series ? what is the main term which determines whether or not the series may

converge ? This is where we need to be able to read mathematically. Consequently, knowing that an

A.P. is a series where we add a constantterm we see that the critical term in the formula is n.


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Then, ifn increases so the sum of the series will increase, and ultimately if so the sum willn d

approach infinity, and the series will diverge : for an A.P. as , so arithmetic series nevern d , Sn d converge.

For a G.P. the situation is different. Some G.P.s converge and some do not. The qualitative analysis is

left as an exercise (see below), but mathematically we have that if thenr < 1

(8)limnd

Sn = limnda(1 rn )

1 r=

a1 r


Interpret the mathematics of the convergence of G.P.s of the textbook. Describe in your own words

what it is about the series, or any part of it, that allows a G.P. to be convergent ? When you have

identified the part(s) of the series which controls convergence, describe the effect on convergence asthis term(s) changes : i.e. what happens to the series if the term(s) is/are small or large ? What happens

to the series as more and more terms are added to the series ? what happens to the series as less and less

terms are added to the series ? Etc....

End of Exercise

1.5 : Arithmetic and geometric means (private reading/self study)

1.5.1: Arithmetic means

Consider three numbers which form an A.P. : . Letp1 be the first number in the series, p2 thep1,p2,p3second number in the series, andp3 be the first number in the series. Then

, , andp1 = a p2 = a + d p3 = a + 2d

Let us calculate the average of these three numbers :

13(p1 +p2 +p3 ) =

13

(a + a + d+ a + 2d)

i.e.13

(p1 +p2 +p3 ) = a + d

But so the mean of three numbers in arithmetic series is given by the middle term.p2 = a + d

Another way of analysing this is by forming certain key combinations of . Then,p1,p2,p3

p1 +p3 = 2a + 2d= 2p2Hence

p2 =12(p1 +p3 )


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Hence we can interpret this as meaning that the average of the first and third terms give the middle

term of the arithmetic series.

1.5.2 : Geometric means

Let us now study the geometric mean of a three numbers in geometric progression. therefore, consider

three numbers which form an G.P. : . Let p1 be the first number in the series, p2 the secondp1,p2,p3number in the series, andp3 be the first number in the series. Then

, , andp1 = a p2 = ar p3 = ar2

Let us calculate the geometric average of these three numbers :

3 p1p2p3 =3 (a.ar.ar2 )

i.e.3 p1p2p3 = ar

But so the geometric mean of three numbers in geometric series is given by the middle term.p2 = ar

Again, another way of analysing this is by forming certain key combinations of . Then,p1,p2,p3

p1.p3 = a2r2 = (p2 )2

Hence

p2 = (p1.p3 )

Hence we can interpret this as meaning that the geometric average of the first and third terms give the

middle term of the geometric series.

Examples

See lecture and refer to Examples 15d, p601 of Bostock and Chandler.

Exercises

Refer to exercises 15d, p603 of Bostock and Chandler.

Exercise on mathematical thinkingExtend the above analysis for arithmetic and geometric means by considering four, five, six, etc...

terms in arithmetic progression and geometric progression :

1) for three terms (as shown above) we saw the arithmetic mean to bep2. Show that for four terms

in arithmetic series the arithmetic mean is , and that for five terms termsp1,p2,p3,p412

(p2 +p3 )

in arithmetic series the arithmetic mean is .p1,p2,p3,p4,p5 p3


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2) for three terms (as shown above) we saw the geometric mean to bep2. Show that for four terms

in geometric series the geometric mean is , and that for five terms inp1,p2,p3,p4 p1.p4 p1,p2,p3,p4,p5geometric series the geometric mean is .p3

Make sure that you can develop the analysis using both of the two different lines of algebra shown in

the sections above

End of Exercise

1.5.3 : The Arithmetic-Geometric mean inequality

Given the arithmetic and geometric means above, we can find a relationship between them as follows.

Consider the difference between two arithmetic terms P1 and P2, Then we have :

P1 P2

Squaring this difference implies that it will be greater than or equal to 0 :

(P1 P2)2 m 0

Therefore

P12 2P1P2 + P2

2 m 0

Adding 4P1P2 to both side we get

P12 2P1P2 + P2

2 + 4P1P2 m 4P1P2

(P1 + P2 )2 m 4P1P2

Implying thatP1 + P2

2m P1P2

Note that the above inequality is true for any numbers x and y and not just for numbers in arithmetic

/geometric sequence. Therefore in general we have that for any two numbers ,a, b c

a + b2

m ab

This is the arithmetic-geometric mean inequality and says that the arithmetic mean is always greater

than or equal to the geometric mean.

We can obtain the same result more directly by using a different binomial term to prove this inequality

: . Then we have

a b2

m 0


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a 2 ab + b m 0so that directly we obtain

a + b2

m ab

Simply from the above example we might predict that

1) 2)a + b + c

3m 3 abc a + b + c + d

4m 4 abcd

for . To prove this let us consider proving 2). As such we might consider doing thea, b, c, dc necessary algebra on . But there is a shorter way of proving 2) if we consider that(a + b + c + d)

4

a + b + c + d4

=12

a + b2

+c + d

2

m a + b2

c + d2

m ab . cd

m 4 abcd

Adopting a similar mathematical trick for the algebra of 1) we can do

a + b + c3

=14 a + b + c +

a + b + c3

m4

abc

a + b + c

3

m (abc)1/4 a + b + c

3

1/4

implying that

a + b + c3

3/4

m (abc)1/4

Taking the 4/3rds power of both sides we finally have

a + b + c3

m 3 abc

Exercise

Prove the following equalities :

1) 2)a + b + c + d+ e

5m 5 abcde a + b + c + d+ e +f

6m 6 abcde


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1.6 : Summation of Number Series

Here we will find the sum of some standard number series. We saw on p4 of these notes that

i=1

n

i =n2

(n + 1)

Lets us now consider the series called the sum of squares :

(1.6.1)r=1

n

r2 = 12 + 22 + 32 + 42 + .... + n2

This is neither an A.P. nor a G.P. so we need to another analysis to find its sum. Consider therefore :

(1.6.2)(r+ 1)3

r3

Expanding we get (r+ 1)3

r3 = r3 + 3r2 + 3r+ 1 r3

Therefore (1.6.3)r=1

n

(r+ 1)3

r3 = 3 r=1

n

r2 + 3 r=1

n

r+r=1

n

1

Writing out the left hand side of (1.5.3) backwards we have

(1.6.4)(n + 1)3

n3 + n3 (n 1)3

+ .... + [33 23 ] + [23 13 ] = 3 r=1

n

r2 + 3 r=1

n

r+r=1

n

1

All the terms of (1.5.4) cancel except the 1st and last terms. Hence

(1.6.5)(n + 1)

3

13

= 3 r=1

n

r2

+ 3 r=1

n

r+ r=1

n

1

Also, we know and also . Hence (1.5.5) becomes r= n/2(n + 1) 1 = n

(n + 1)3

13 = 3 r=1

n

r2 + 3(n2

(n + 1)) + n

From which (after some algebra) we can find to be r2

(1.6.6)r=1

n

r2 =n6

(n + 1)(2n + 1)


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Interpret the mathematics of the proofs above. Describe in your own words what it is about doing

which allows the proof to work. Describe any and all steps in the proof(r+ 1)3

r3

End of Exercise


1) In the proof above we used the identity . Would we get the same result if we used(r+ 1)3

r3

the identity ? If so, why does the identity work ? If not why does the identity not work ?r3 (r 1)3

2) Use the identity to prove that(r+ 1)2

r2 r=1

n

r=n

2

(n + 1)

3) Find a relevant identity which would allow you to prove that r=1

n

r3 =n2

4(n + 1)

2

4) Find a proof for finding the sum : by using a relevant identityr=1

n

r4

5) From the results of , , and can you predict the formula for the sums r r2 r3 r4 rn, n m 5

End of Exercise

Examples

See lecture

Exercises

See exercise sheets

1.7 : Method of Induction

So far we have looked at series from the perspective of finding its sum by using terms and

differences/ratios of the series. Here we will consider whether or not the general formula for such sums

are in fact true or not. We will therefore use a method of proof known as Proof by Induction to showwhether or not these formula are correct.

As an example, consider the following series :

(1.7.1)Sn = r=1

n

r.(r+ 1) = (1)(2) + (2)(3) + (3)(4) + .... + (n)(n + 1)


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This series does not fit the normal A.P. or G.P. structure so we cant use their formulae to find the sum

of this series . However if we consider the partial sums of (1.7.1) :

S1 = (1)(2) = 2 =13

(1)(2)(3)

S2 = (1)(2) + (2)(3) = 8 =13

(2)(3)(4)

S3 = (1)(2) + (2)(3) + (3)(4) = 20 =1

3 (3)(4)(5)

we can deduce the general pattern for the sum of the series to be

(1.7.2)Sn =13 n.(n + 1)(n + 2)

However, we now need to prove that (1.7.2) is true for all n and not just for a few values. We do this in

two steps :

firstly we need to prove that (1.7.2) actually works for at least one number. Usually we

take the easiest number to use, i.e. 1 .....

..... then we need to prove that if the formula is true for any value k, then it is true for

any value k+1

Hence for (1.7.2) :

the 1st stage :

Here we test (1.7.2) for a specific value ofn, say n = 1, and compare against the series, to see if

the formula is valid. So,

for n = 1, Sn =13 .1.(2)(3) = 2

And (1.7.1) gives, for n = 1,

Sn = r=1

1

r.(r+ 1) = 1.(1 + 1) = 2

So we know that formula (1.7.2) does work for n = 1.

the 2nd stage :

The second stage in our proof is in two steps. The 1st step is a very formal step. Specifically let

the formula be true for any number n = k, . Then we have1 < k< n

(1.7.3)Sk =13 .k.(k+ 1)(k+ 2)

The 2nd step is to now show that if we add k+1 to both (1.7.1) and (1.7.2) they will both be

equal, in other words the answer we get from (1.7.2) will be the same we get from (1.7.1)

(which we know is the true answer) when we add the (k+1)th term.


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Hence we need to show that

(1)(2) + (2)(3) + .... + (k)(k+ 1) + (k+ 1)(k+ 2) = 13 .k.(k+ 1)(k+ 2) + (k+ 1)(k+ 2)

(1.7.4)

The left hand side of (1.7.4) is simply (1.7.1) with n = k+1, i.e. . We thereforer=1k+1 r.(r+ 1)

need to show that the right hand side simplifies to (1.7.3) with k replaced by k+1. Simple

algebra shows that for the right hand side of (1.7.4) we get

13 .k.(k+ 1)(k+ 2) + (k+ 1)(k+ 2) =

13(k+ 1)(k+ 2)(k+ 3)

Hence (1.7.4) becomes

(1)(2) + (2)(3) + .... + (k)(k+ 1) + (k+ 1)(k+ 2) =13(k+ 1)(k+ 2)(k+ 3)

i.e. (1.7.5)r=1

k+1

r.(r+ 1) = (1)(2) + (2)(3) + .... + 13 (k+ 1)(k+ 2)(k+ 3)

which is what we wanted to obtain. Hence we have proved that (1.7.2) is true

We can therefore state this process of proof by induction more formally as :

Let S(n) be a series for integers n, and let the

following two statements be true :

i) S(1) is true

ii) for all , if S(k) is true then S(k+1) iskm 1also true

Then for all , S(n) is truen m 1

Examples

Prove that

1) 2)1 + 3 + 5 + .... + (2n 1) = n2 1 + 2 + 3 + 4 + ... + n =12 n(n + 1)

See lecture

Exercises

See exeercise sheet


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Interpret the mathematical process of doing a proof by induction. Describe, in your own words, all the

steps done in the example above. What is the final aim we want to achieve ? How do we go about

achieving that aim mathematically ? Explain the reason for adding the (k+1)th term. Why do we not

need to add the (k+2)th term or any other term ?

End of Exercise

1.8 : Test for Convergence of Series

Let us now study more deeply how to analyse the convergence of a series. Series can do one of three

types of things : They can converge, they can diverge, or they can neither converge nor diverge.

consider then the series

1 + 12 + (12 )

2 + ( 12 )3 + ( 12 )

4 + .....

has the sum to one term (S1) of 1 ; has the sum to two term (S2) of 1.5

has the sum to three term (S3) of 1.75 ; has the sum to four term (S4) of 1.875

has the sum to five term (S5) of 1.9375

It seems therefore that the sum to infinity approach 2. If we look at the n th term we see that

Sn = 2 2(12

)n

and as , , and therefore . This series is therefore convergent.n d (1

2

)n d 0 Sn d 2

If we now look at the arithmetic series

1 + 2 + 3 + 4 + 5 + .....

we see that S1 = 1, S2 = 3, S3 = 6, ....., implying that as . This series isSn =n2(n + 1) n d , Sn d

divergent.

Finally the series

1 1 + 1 1 + 1 1 + ....

neither converges nor diverges because the sum of more and more terms keeps changing/alternating

between 0 and 1. On this course we will only ever consider series which have positive terms.


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1.8.1 : Test 1 : Limit test

Let us consider the series

Sn = u1 + u2 + u3 + ..... + un + .....

If the series converges then as ,n d Sn d k

This implies that the difference between any two sums becomes smaller and smaller, i.e. that

as .Sn Sn1 d 0 n d

But . Hence as n gets larger and larger the general term un of the series must approach 0,Sn Sn1 = uni.e.

limnd

un = 0

A general definition of convergence can now be stated :

is convergent ==> (1.8.1)un limnd un = 0

Note however that =/=> is convergent (1.8.2)limnd

un = 0 un

i.e. it will not always be the case that if then the series will converge. In fact nothing canlimnd un = 0be said about the convergence properties of the series. This therefore means that the condition in

(1.8.1) can only be used to test that a series is not convergent, i.e.

if then is divergent (1.8.3)limnd

un ! 0 un

Examples

According to the limit test, the series

1 +13

+19

+181

.....

converges, but the series

1 + 3 + 9 + 81 + ....

obviously does not converge

1.8.2 : Proof of limit test

We want to prove that if then is divergent. To do this we will try to prove the opposite,limnd

un ! 0 unnamely that

if converges then . un limnd un = 0


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To do this let be the sum of the first kterms. Hence, the kth term uk can be found from the partialSksums as

uk = Sk Sk1Taking limits we have

limkd

uk = limkd

(Sk Sk1 )

= limkd

Sk limkd

Sk1

But as kt the partial sums become the total sum S. Therefore,

limkd

uk = S S = 0

Therefore, since we have proved that if the series converges then , this means that iflimnd

un = 0

then the series diverges. QEDlimnd

un ! 0

ExamplesSee lecture

1.8.3 : Test 2 : Comparison test

Consider the series below and use the limit test above to see if the series below converges or not :

(*)Sn = 1 +12 +

13 +

14 +

15 + .....

This series is called the Harmonic series. Let us group the terms of this series as follows :

(1.8.4)1 + 12 + ( 13 + 14 ) + ( 15 + 16 + 17 + 18 ) + ( 19 + .... + 116 ) + ....

Make all the terms in each bracket the same as the last term of each bracket, i.e.

(1.8.5)1 +12 + (

14 +

14

) + (18 +

18 +

18 +

18

) + (1

16 + .... +1

16) + ....

Comparing the sum of terms between (1.8.4) and (1.8.5), we see that the sum of the terms in each

bracket of (1.8.5) is less than the sum of the terms in each bracket of (1.8.4). But (1.8.5) can be

re-written as

(1.8.6)Sn = 1 +

12 + (

12

) + (12

) + (12

) + .....

Q : Does series (1.8.6) converge or diverge ? Well, it diverges, i.e.

as n d , Snd


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But we saw that Sn of (1.8.4) was greater than . So, if diverges, then the original series Sn alsoSn Sn

diverges, even though in (1.8.4) !limnd

un = 0

Exercise on thinking mathematically

Describe, in your own words, the mathematical idea behind the development of expressions (*) to

(1.8.6) above. What is it about what we have done that makes the whole idea work ? Why do we want

to bracket the terms in the way we have done it in (1.8.4) and (1.8.5) ? How is this useful ? How does

this allows us to study the convergence/divergence characteristics of (*) ? What is it about (1.8.5) that

allows us to say that (*) is divergent ?

End of Exercise

Exercise on thinking mathematically1) Compared with the grouping of terms in (1.6.4) why is it not useful for us to consider the

grouping below ? :

(1.8.7)1 + (12 +

13

) + (14 +

15 +

16 +

17

) + (18 +

19 + .... +

115

) + ....

(1.8.8)1 +12 +

13 + (

14 +

15

) + (16 +

17 +

18 +

19

) + (1

10 + .... +1

17) + ....

Describe both informally (i.e. in English description) and formally (i.e. in mathematical

notation) what happens to convergence/divergence aspects when we consider the grouping of (1.8.7)

and (1.8.8).

Note that if you truly understand the effect of the grouping of (1.8.4) and why it was done that

way then you will see that there is a difference in the effect of the groupings of (1.8.7) and (1.8.8).

Describe, both informally and formally, and in detail, the differences in the effects of the groupings of

(1.8.7) and (1.8.8). What does this imply about the way you have to group terms in the series ?

2) Difficult ? : Having identifies the true nature of the way we must group terms, create/find other

series which you can prove are divergent but which are based on grouping terms only two at a time,

three at a time, or four at a time, i.e.

or1 + ( ## + ## ) + ( ## + ## ) + ( ## + ## ) + ..... 1 + ( ## + ## + ## ) + ( ## + ## + ## ) + .....

etc... This means you will have to experiment with the structure and the terms of series so that you can

develop one which has the necessary structure to prove that it is divergent.

End of Exercise


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The process for testing that the Harmonic series was divergent is called the comparison test for

convergence. The comparison test therefore relies on comparing the terms of the series with the

(smaller) terms of another series we know diverges. More generally we may then say that

if all the terms of a given series are greater than the terms of a

series we know diverges, then that original series must also divergeSimilarly

if all the terms of a given series are less than the terms of a series

we know converges, then that original series must also converge

The comparison test only works for series containing positive terms. Mathematically we may then

make the following definition : provided andun > 0 an > 0

1) if , whereNis fixedun > an n >N

is divergent ==> is divergent ar ur

2) if , whereNis fixedun < an n >N

is convergent ==> is convergent ar ur


Interpret the mathematical definition above Describe, in your own words, what each part(s) of the

definition means and what it implies. what do symbols and groups of symbols mean ? What are the

effects of the relevant operations ?

End of Exercise

1.8.4 : Proof of the comparison test

Let be the series we are testing and let be the series we know converges. Then uk ak

a1 + a2 + a3 + .... + ak+ .... [A

whereA represents the sum of the series . Since then

ak

u1

< a1, u

2< a

2, u

3< a

3,....u

k< a

k,....

u1 + u2 + u3 + .... + uk+ .... < a1 + a2 + a3 + .... + ak+ .... [A

hence converges. Generally speaking, if the bigger series converges, then the smaller uk akseries also converges. To prove divergence, we can use the concept opposite to the above : in ukshort if we know , then since , also. akd uk > ak akd


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Then, to use the comparison test we need :

1) an expression for the general term of the series

2) a series to compare with which we know is either convergent or divergent. Some useful

series are :

Convergent series : n=1

1n

k

n=0

1n! n=1

1n(n + 1)

Divergent series : n=1

1n

Other know tests can be found in the extra handouts.

Examples

See lecture.


You might think that series of the form should either all diverges or all converge. But this is not 1pthe case. Some will converge and some will diverge.

So what is it about the nature or character of when (i.e. and ) which 1p p = n orp = n! 1n!

1n

makes the difference ? What is it about the fractions and/or the sum of fraction terms which makes one

type of series converge and the other type of series diverge ?

End of Exercise

Let us examine the harmonic series in more detail. We want to study its convergence properties and

prove when it is convergent and when it is divergent. We will discuss this in the lecture, but in

summary we have that :

diverges whenn=1

1n

k

k[ 1

converges whenn=1

1n

kk> 1

The proof of these two situations is in the extra handouts (ignore the small section on : Another form

of the comparison test).


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1.8.5 : Test 3 : Ratio test

We now move to another way of testing the convergence or not of a series. This is similar to one form

of the comparison test we did in the examples above. There we could determine convergence by doing

either or as . However, in the ratio test we test , i.e.un/an > 1 un/an < 1 n d un+1/un

if is a series of positive terms : un

if the series convergeslimnd

un+1un < 1

if the series divergeslimnd

un+1un > 1

if we cannot decide if the series converges or divergeslimnd

un+1un = 1

We will not look at the proof since this is beyond the scope of our course.

Examples

Do the following series converge or diverge ?

i) ii)1 +32

+522

+723

+ .....12

+23

+34

+45

+ .....

See lecture.

Exercises

See exercise sheets


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Series : Extra Topics

1.0 : IntroductionThis is an extra set of notes which is included free (!) if you are interested in reading them. This extra

set of notes is not part of the syllabus, and you may therefore leave it aside if you wish. However, if

you are interested in reading on, then you will read about the topic of sequences and their properties.

This will include :

Introduction to Sequences

Here we will define what sequences and define the notation used in the study of these.

Recurrence Relations

Here we will look at types of formula which give rise to sequences. These formula contain

terms which refer to previous elements of a given sequences. As such they therefore allow us todevelop infinite sequences from simple expressions.

Fibonacci Sequences

Here we introduce the most well known of sequences, and study some basic numerical patterns

and properties of this sequence.

--------------------------------------------------------------------------------------------------

1.1 : Sequences

Here we will summarise what sequences and series are, and the notation used in the study of these.

1.1.1 : Introduction

Consider, therefore, the following set of numbers :

1) 2, 3, 4, 5, ....... 2) 2, 4, 6, 8, 10, ...... 3) 4, 9, 16, 25, ......

Informally we may say that : in 1) each number is one more than the previous number

in 2) each number is twice the previous number

in 3) each number is the square of the natural numbers 1, 2, 3, ....

Formally we may then define a sequence as :

a set of terms (numbers, variables, etc...) in a specific

order with a rule for obtaining these terms


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1.1.2 : Stating sequences mathematically

To describe a sequence mathematically we need three things :

1) a symbol which refers to the number of the sequences

2) a subscript symbol which refers to the position of the number in the sequence

3) a rule for calculating how each term of the sequence is calculated

As an example we can then write the sequence of natural numbers

1, 2, 3, 4, 5, 6, .......

as a1, a2, a3, a4, a5, a6,....

Therefore the complete mathematical statement representing the sequence of natural number is

ai, i = 1,2,3,.....|an = n

Exercise

1) Develop mathematical statements representing the following sequence

i) 2, 4, 6, 8, ..... ii) 1, 3, 5, 7, .....

iii) 1, 4, 9, 16, 25, ..... iv) 2, 4, 8, 16, 32, 64, 128, .....

v) -1, 1, -1, 1, -1, 1, .....

2) Use your answer to 1) v) above to develop a mathematical statement representing the following

sequence :

1, 1, 3, 3, 5, 5, 7, 7, 9, 9, .....

ans : an=n-(1+(-1)n

)/2

3) Given the sequence and your answer to 1) v) above develop aa i, i = 1,2,3, .....|an = 1

mathematical statement representing the following sequence :

0, 2, 0, 2, 0, 2, 0, 2, ......

ans : add the respective terms of the sequences

1.2 : Recurrence Relations

Important in the sequences we will study later is the concept of a recurrence formula, or recurrence

relations. A recurrence relation for the sequence is an equation that expresses an in terms of oneanor more of the previous terms of the sequence, namely, a0, a1 , , an-1, for all integers . Thisn m 0recurrence formula is then used to generate the numbers of the sequence, and as such the sequence is a

solution to the recurrence relation.


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Example 1

Consider the recurrence relation

a i, i = 2,3,4,...|an = 2an1 an2

In order to be able to generate a sequence from this recurrence relation we need to provide initialstarting values, and the number of starting values depends upon how many terms there are in the right

hand side of the recurrence relations. In this case we need to define two starting values a0 and a1. Hence

let . then we have the sequence :a0 = 0 and a1 = 1

0, 1, 2, 3, 4, 5, 6, ......

Example 2

Consider the previous recurrence relation . Are the sequences givenai, i = 2,3,4,...|an = 2an1 an2by

i) and ii)an = 3n an = 5

solutions of this recurrence relation ? To show this we can substitute each of the suggested sequences

into the recurrence relation to see if they make these work :

for :an = 3n 2an1 an2 = 2(3(n 1)) 3(n 2) = 3n = an

for :an = 5 2an1 an2 = 2(5) 5 = 5 = an

Therefore, both sequences of are solutions to the recurrence relation.an

Exercises

1) Show that the sequence satisfies the recurrence relation1,1!,2!, 3!, 4!, ..., (1)nn! for n m 0

forSk= k.Sk1 km 1example 8.1.4, p426, Epp

(*all the following are from No 9 onwards, p438, Epp*)

2) Show that the sequence given by , for , satisfies the recurrence relationan = 3n + 1 n m 0

forak= ak1 + 3 km 1

3) Show that the sequence given by , for , satisfies the recurrence relationbn = 5n n m 0

forbk= 5bk1 km 1

4) Show that the sequence 0, 1, 3, 7, ..., , for , satisfies the recurrence relation2n 1 n m 0

forck= 2ck1 + 1 km 1


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5) Show that the sequence 2, 3, 4, 5, ..., 2 + n, for , satisfies the recurrence relationn m 0

fortk= 2tk1 tk2 km 2

6) Show that the sequence 0, 1, 3, 7, ..., , for , satisfies the recurrence relation2n

1 n m 0

fordk= 3dk1 2dk2 km 2

There are many different types of sequences, some of which give rise to interesting mathematical

patterns we shall now study.

1.3 : Fibonacci Sequences

This is probably the most well know sequence and was discovered through a mathematics competition

in 1225. The problem for this competition was based finding out how fast rabbits would breed in idealcircumstances. This was assuming that none of the rabbits died, and that each pair of rabbits did not

give birth to more than one pair of rabbits each month.

The picture below shows a pair of rabbits that are born and only breed after the first month. After the

first month they give birth to a pair of rabbits which themselves only breed after their first month. This

pattern is then repeatedly continuously and can be seen in the diagram below

When we look at the sequence and pick a number we notice that it is composed of the previous two

numbers added together. 1 + 1 = 2 ... 1 + 2 = 3... 2 + 3 = 5 and so on.


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The sequence then tends to be written as

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, ...........

and the recurrence relation which satisfies this sequence is Fibonacci recurrence relation

Fn = Fn1 + Fn2

and this simply allows us to generate the next number in the sequence by adding the previous two

numbers in the sequence.

Exercises

1) Develop an expression for Fn in terms of

i) Fn-2 and Fn-3 only ; ii) Fn-3 and Fn-4 only

iii) Fn-4 and Fn-5 only ; iv) F1 and F0 only

(*all the following are from No 24 onwards, p438, Epp*)

2) Show, or prove by induction, that , for allFk2 Fk1

2 = Fk.Fk+1 Fk+1.Fk1 km 1

3) Show, or prove by induction, that , for allFk+12 Fk

2 Fk12 = 2Fk.Fk1 km 1

4) Show, or prove by induction, that , for allFk+12 Fk

2 = Fk1.Fk+2 km 1

1.3.1 : Divisibility of sets of Fibonacci numbers

The Fibonacci sequence has some interesting arithmetic properties. Specifically, sets of Fibonaccinumbers have certain divisibility properties :

Divisibility by 11

The sum of any ten consecutive Fibonacci numbers is always divisible by 11 and gives a

answer which is itself a Fibonacci number :

17567/11 = 1579979/11 = 89143/11 = 13

6,76537755

4,18123334

2,58414421

1,5978913

987558610345

377213

233132

14481

8951


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Divisibility by Fibonacci numbers

As the consecutive Fibonacci integers (Fn) increase note how they are divisible by consecutive

Fibonacci numbers :

Every 3rd Fibonacci number is divisible by 2.

Every 4th Fibonacci number is divisible by 3.Every 5th Fibonacci number is divisible by 5.

Every 6th Fibonacci number is divisible by 8.



etc....


1) What happens if you double/triple/... every Fibonacci number ? Is there another Fibonacci

sequence within this ? ...

2) ... then what about the divisibility properties of the sequences you have developed in 1) ? Are

their any numbers in your sequences of 1) wich are divisible by any other numbers in the

sequence ?

End of Exercise

Pascals Triangle and the Fibonacci sequence

Consider Pascals triangle. The diagram below shows that we can sum the numbers of the

diagonal to obtain the Fibonacci sequence :

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5

1 6 15 20 15 6 1

1

11235813Fn:


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1) How far does this diagonal property work ? If it breaks down at some specific diagonal, devise

a way of fixing it (in any way you can make work) so that this property works again ?

End of Exercise

1.3.2 : The golden ratio

Another important number that can be found as a result of the Fibonacci sequence is 1.61803. This

number is called (Phi) which is the Golden Ratio and is the limit of the ratio of two consecutive

Fibonacci numbers, i.e.

1/1 = 1 ---> 1/2=0.5 ---> 2/3=0.66666666

3/5=0.6 ---> 5/8=0.625 ---> 8/13=0.6153846

13/21=0.619047 ---> 21/34=0.617647 ---> 34//55=0.6181818

and ultimately 1618033988749894.... A graph of the convergence to can = limnd F(n + 1)/F(n) =be seen below :

The golden ratio can in fact be obtain by solving the quadratic . To see this consider anyx2 x 1 = 0three Fibonacci numbers . Then, for very large values of i, the ratio ofF(i)F(i), F(i + 1), and F(i + 2)

and F(i+1) will be almost the same as the ratio F(i+1) and F(i+2) (this is a standard number property :

i.e. 99/98 is closer to 98/97 than 2/3 is to 3/4)


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So let's see what happens if both of these ratios have the same value x :

F(i + 1)

F(i)=

F(i + 2)

F(i + 1)=x

Substituting into the above we have :F(i + 2) = F(i + 1) + F(i)

F(i + 1)

F(i)=

F(i + 1) + F(i)

F(i + 1)=x

F(i + 1)

F(i)= 1 +

F(i)

F(i + 1)=x

F(i + 1)

F(i)= 1 +

1x =x

hence x2 =x + 1

the solution of which is the golden ratio . Since is a root to this quadratic then

2 = + 1

and what this tells us is that is a number such that, in order to square it, all we need to do is to add 1

to it. Solving this quadratic we get

and = 12 +5

2 =12

5

2

namely, = 16180339887... and = 06180339887... The root is then the golden ratio number .However, there are some interesting properties of the golden ratio (which are actually connections

between and ) :

0) squaring property : 2 = + 1 (as above)

1) reciprocal property : = -1/

2) unitary property : . = -1

3) sum property : + = 1

4) difference property : = 5

1.4 : Mathematical Analysis of Fibonacci SequencesHere we will analyse certain mathematical properties of the Fibonacci sequence.

1.4.1 : Immediate successors and predecessors of Fibonacci numbers

This analysis is taken from Note 86.60 of Mathematical Gazette 2001. We know that in order to

calculate a Fibonacci number from the Fibonacci sequence :

;Fn = Fn1 + Fn2 F1 = F2 = 1, n m 2


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we need to know the previous two numbers in the sequences. Here we shall show that this is in fact not

the case. We will see that it is possible to calculate the next Fibonacci number knowing only the

immediately previous Fibonacci number.

To see this let and . Let us now start with the difference = (1 + 5 )/2 = (1 5 )/2 = 1 = 1/

between two Fibonacci numbers as :

Fn+1 Fn

Using the standard recurrence relation for the Fibonacci sequence we have

i)Fn+1 Fn = (Fn + Fn1 ) Fn = (1 )Fn + Fn1

ii)= Fn + Fn1

Factorising in this last expression we have

iii)= (Fn Fn1 )

Comparing iii) with the left hand side of i) we see that they have the same structure, i.e. they have an

structure. This means that the analysis which allowed us to get from the LHS of i) to iii) canF... F...be used on iii). Hence

Fn+1 Fn = (Fn Fn1 )

= 2(Fn1 Fn2 )

= 3(Fn2 Fn3 )

..............................

= n1(F2 F1 ) = n1(1 ) = n

However, we know that hence for . This implies that < 0.62 2

Topic4 Series

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Transcript of Topic4 Series