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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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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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
Exercise on reading mathematically
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.
Exercise on reading mathematically
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
Exercise on mathematical thinking
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
Exercise on reading mathematically
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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Exercise on reading mathematically
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
Exercise on mathematical thinking
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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Exercise on reading mathematically
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
Exercise on reading mathematically
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.
Exercise on thinking mathematically
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.
Every 7th Fibonacci number is divisible by 13.
Every 8th Fibonacci number is divisible by 21.
etc....
Exercise on thinking mathematically
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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Exercise on thinking mathematically
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