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MATHEMATICAL ANALYSIS A STRAIGHTFORWARD
APPROACH
K.G. BINMORE Professor of Mathematics L.S.E.
.
CAMBRIDGE

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MATHEMATICAL ANALYSIS A STRAIGHTFORWARD
APPROACH
K.G. BINMORE Professor of Mathematics L.S.E.
.
CAMBRIDGE

...
UNIVERSITY
...
 ..~.: .
PRESS
CAMBRIDGE LONDON
 NEW YORK
 MELBOURNE
Published by the Syndics of the Cambridge University Press The Pitt Building, Trumpington Street, Cambridge CB2 1RP Bentley House, 200 Euston Road, London NW1 2DB 32 East 57th Street, New York, NY 10022, USA 296 Beaconsfield Parade, Middle Park, Melbourne 3206, Australia © Cambridge University Press 1977
First published 1977 Reprinted 1978 Printed in Great Britain at the University Press, Cambridge Library of Congress Cataloguing in Publication
Binmore, K G 1940Mathematical analysis. Includes index. 1. Mathematical analysis. I. Title. QA300.B536 515 7628006 ISBN 0 521 214807 hard covers ISBN 0 521 291674 paperback
Data
CONTENTS
Preface
ix
1
Real numbers
1.1 1.2 1.3 1.4 1.9 1.10 1.13 1.14
Set notation The set of real numbers Arithmetic Inequalities Roots Quadratic equations Irrational numbers Modulus
1 2 3 3 6 6 9 10
2
Continuum Property
12
2.1 2.2 2.6 2.7 2.9 2.11
Achilles and the tortoise The continuum property Supremum and infimum Maximum and minimum Intervals Manipulations with sup and inf
12 13 15 15 16 18
3
Natural numbers
3.1 3.2 3.7
Introduction Archimedean property Principle of induction
20 20 22
4
Convergent sequences
26
4.1 4.2 4.4 4.7 4.15 4.21 4.26
The bulldozers and the bee Sequences Definition of convergence Criteria for convergence Monotone sequences Some simple properties of convergent sequences Divergent sequences
26 27 27 30 33 37 38
5 5.1 5.8
Subsequences
41
Subsequences Bo1zano Weierstrass theorem
41 46
1
20
V
vt
Contents
5.12 5.16
Lim sup and lim inf Cauchy sequences
47 49
6
Series
53
6.1 6.4 6.7 6.12 6.20 6.23
Definitions Series of positive terms Elementary properties of series Series and Cauchy sequences Absolute and conditional convergence Manipulations with series
53 54 57 60 61
7 7.1 7.6 7.9 7.11 7.13
Functions
64
Notation Polynomial and rational functions Combining functions Inverse functions Bounded functions
64 66 68 68 70
8
Limits of functions
74
8.1 8.2 8.3 8.6 8.8 8.11 8.16 8.18
Limits from the left Limits from the right [(x) ) I as x ) ~ Continuity at a point Connexion with convergent sequences Properties of limits Limits of composite functions Divergence
74 74 75 77 78 79 81 82
9
Continuity
84
9.1 9.7
Continuity on an interval Continuity property
84 86
10
Differentiation Derivatives Higher derivatives More notation Properties of differentiable functions Composite functions
91 91 92 93 95 98
10.1 10.2 10.4 10.5 10.12 11 11.1 11.3 11.5 11.9
56
Mean value theorems
100
Local maxima and minima Stationary points Mean value theorem Taylor's theorem
100 101 102 105
Contents
vii
12
Monotone functions
108
12.1 12.3 12.6 12.9 12.11 12.13
Definitions Limits of monotone functions Differentiable monotone functions Inverse functions Roots Convex functions
108 108 110 110 112 114
13 13.1 13.2 13.3 13.9 13.16 13.19 13.27 13.31
Integration
119
Area The integral Some properties of the integral Differentiation and integration Riemann integral More properties of the integral Improper integrals EulerMaclaurin summation formula
119 120 121 124 127 129 132 134
14
Exponential and logarithm
137
14.1 14.4 14.6
Logarithm Exponential Powers
137 140 141
15
Power series
143
15.1 15.4 15.7
Interval of convergence Taylor series Continuity and differentiation
143 145 147
16
Trigonometric functions
150
16.1 16.2 16.4
Introduction . Sine and cosine Periodicity
17
The gamma function
17.1 17.3 17.5
Stirling's formula The gamma function Properties of the gamma function
18
Appendix
150 151 153 156 156 157 158
161 This contains the proofs of 'propositions' left unproved in the main body of the text.
Solutions to exercises
172
Suggested further reading
251
Notation
253
Index
255
PREFACE
This book is intended as an easy and unfussy introduction to mathematical analysis. Little formal reliance is made on the reader's previous mathematical background, but those with no training at all in the elementary techniques of calculus would do better to turn to some other book. An effort has been made to lay bare the bones of the theory by eliminating as much unnecessary detail as is feasible. To achieve this end and to ensure that all results can be readily illustrated with concrete examples, the book deals only with 'bread and butter' analysis on the real line, the temptation to discuss generalisations in more abstract spaces having been reluctantly suppressed. However, the need to prepare the way for these generalisations has been kept well in mind. It is vital to adopt a ~stematic approach when studying mathematical analysis. In particular, one should always be aware at any stage of what may be assumed and what has to be proved. Otherwise confusion is inevitable. For this reason, the early chapters go rather slowly and contain a considerable amount of material with which many readers may already be familiar. To neglect these chapters would, however, be unwise. The exercises should be regarded as an integral part of the book. There is a great deal more to be learned from attempting the exercises than can be obtained from a passive reading of the text. This is particularly the case when, as may frequently happen, the attempt to solve a problem is unsuccessful and it is necessary to turn to the solutions provided at the end of the book. To help those with insufficient time at their disposal to attempt all the exercises, the less vital exercises nave been marked with the symbol t. (The same notation has been used to mark one or two passages in the text which can be omitted without great loss at a first reading.) The symbol * has been used to mark exercises which are more demanding than most but which are well worth attempting. The final few chapters contain very little theory compared with the number of exercises set. These exercises are intended to illustrate the power of the techniques introduced earlier in the book and to provide the opportunity of some revision of these ideas. This book arises from a course of lectures in analysis which is given at the London School of Economics. The students who attend this course are mostly not specialist mathematicians and there is little uniformity in their previous
ix
Preface
x
mathematical training. They are, however, quite wellmotivated. The course is a 'one unit' course of approximately forty lectures supplemented by twenty informal problem classes. I have found it possible to cover the material of this book in some thirty lectures. Time is then left for some discussion of point set topology in simple spaces. The content of the book provides an ample source of examples for this purpose while the more general theorems serve as reinforcement for the theorems of the text. Other teachers may prefer to go through the material of the book at a more leisurely pace or else to move on to a different topic. An obvious candidate for further discussion is the algebraic foundation of the real number system and the proof of the Continuum Property. Other alternatives are partial differentiation, the complex number system or even Lebesgue measure on the line. Iwould like to express my gratitude to Elizabeth Boardman and Richard Holmes for reading the text for me so carefully. My thanks are also due to 'Buffy' Fennelly for her patience and accuracy in preparing the typescript. Finally, Iwould like to mention M.C. Austin and H. Kestelman from whom Ilearned so much of what Iknow. July 1976
K.G.B.
I
REAL NUMBERS
1.1
Set notation
A set is a collection of objects which are called its elements. If x is an element of the set S, we say that x belongs to S and write xES.
If y does not belong to S, we write y t/=. S. The simplest way of specifying a set is by listing its elements. We use the notation A =
U, 1,y2, e, 1T}
to denote the set whose elements are the real numbers ~, I, Similarly
y2, e and
1T.
B = {Romeo, Juliet} denotes the set whose elements are Romeo and Juliet. This notation is, of course, no use in specifying a set which has an infinite number of elements. Such sets may be specified by naming the property which distinguishes elements of the set from objects which are not in the set. For example, the notation
c
=
{x: x
> O}
(which should be read 'the set of all x such that positive real numbers. Similarly
x> 0') denotes the set of all
D = {y: y loves Romeo} denotes the set of all people who love Romeo. It is convenient to have a notation for the empty set 0. This is the set which has no elements. For example, if x denotes a variable which ranges over the set of all real numbers, then
{x :x2 + 1 = O} = 0. This is because there are no real numbers x such that x2 = 1. If Sand T are two sets, we say that S is a subset of T and write
1
2
Real numbers SC T
, if every element of S is also an element of T. As an example, consider the setsP = {I, 2, 3, 4} and Q = {2, 4}. Then Q C P. Note that this is not the same thing as writing Q E P, which means that Q is an element of P. The elements of P are simply 1,2,3 and 4. But Q is not one of
these. The sets A, B, C and D given above also provide some examples. We have Ace and (presumably) BCD.
1.2
The set of real numbers
It will be adequate for these notes to think of the real numbers as being points along a straight line which extends indefinitely in both directions. The line may then be regarded as an ideal ruler with which we may measure the lengths of line segments in Euclidean geometry.
1
2
I
.1
I
o
2
I
,j2 2
4
3
The set of all real numbers will be denoted by IR. The table below distinguishes three important subsets of IR.
Subset
Notation
Elements
Natural numbers (or whole numbers)
N
1,2,3,4,5,
71..
.. 2,1,0,1,2,3,
Integers
... ...
Rational numbers (or fractions)
Not all real numbers are rational. Some examples of irrational numbers are
y2, e and
1T.
While we do not go back to first principles in these notes, the treatment will be rigorous in so far as it goes. It is therefore important to be clear, at every stage, about what our assumptions are. We shall then know what has to be proved and what may be taken for granted. Our most vital assumptions are concerned with the properties of the real number system. The rest of this chapter and the following two chapters are consequently devoted to a description of the
Real numbers
3
properties of the real number system which we propose to assume and to some of their immediate consequences.
1.3
Arithmetic
The first assumption is that the real numbers satisfy all the usual laws of addition, subtraction, multiplication and division. The rules of arithmetic, of course, include the proviso that division by zero is not allowed. Thus, for example, the expression 2
o makes no sense at all. In particular, it is not true that
o2 = .
,
00
We shall have a great deal of use for the symbol but it must clearly be does not represent a real number. Nor can it be treated as understood that such except in very special circumstances. 00,
00
1.4
Inequalities
The next assumptions concern inequalities between real numbers and their manipulation. We assume that, given any two real numbers a and b, there are three mutually exclusive possibilities:
(i)
a> b
(ii) a = b
(a is greater than b) (a equals b)
< b (a is less than b). Observe that a < b means the same thing (iii) a
as b
> a. We:have,
for example, the
following inequalities. 1>0; 3>2; 22; 1>1; 2~3; 1~0;
3~?
We assume four basic rules for the manipulation of inequalities. From these
4
Real numbers
the other rules may be deduced. (I) If a > band b > e, then a > e.
I
I
c
b
I a .
(II) If a> band c is any real number, then a+e>b+e. (III) If a> band e > 0, then ae through by a positive factor). (IV) If a> band e < 0, then ae reverses the inequality).
1.5
Example
> be (i.e. inequalities < be
can be multiplied
(i.e. multiplication by a negative factor
If a> 0, prove that aI> O.
Proof We argue by contradiction. Suppose that a > 0 but that aI ~ O. ~t cannot be true that aI = 0 (since then 0 = O~a = 1). Hence
aI
< O.
By rule III we can multiply this inequality through by a (since a 1 = al.a 0). Hence
= O.
Therefore the assumption aI ~ 0 was false. Hence
aI> O.
1.6
If x andy are positive, then x O.
(1)
I'
t
Real numbers
I
t
I
5
Since x + y > 0, (x + yfl through inequality (1) by (x
> 0 (example 1.6). We can therefore + y rl to obtain
multiply
yx>O
x 0, a
Example
a b. Then a  b > O. But, for any e> 0, particular case when e = b  a. Thus
+ e. Hence a < b + e in the a
and so
1.8
I t i.~:
Exercise
(1) If x is any real number, prove that x2 prove that (i) 0 (2) If b
a
b > 1.
> 0 and B > 0 and A
b band c > d, prove that a + c > b + d (Le.lnequalities can be added). If, also, b > 0 and d > O,prove that ac > bd (Le. inequalities between ~umbers can be m~ltiplied). (4) Show that each of the following i~s may fail to hold even though a > band c > d.
(i) a c >bd
6
Real numbers
.. a (11) ~
>d
(iii) ac
> bd.
b
What happens if we impose the extra condition that b > 0 and d > O? (5) Suppose that, for any E > 0, a  E < b < a + E. Prove that a = b. (6) Suppose that a < b. Show that there exists a real number x satisfying a c. x c.b,
1.9
Roots
Let n be a natural number. The reader will be familiar with the notation y =xn. For example.x" =x.x and x? =x.x.x. Our next assumption about the real number system is the following. Given any y ~ 0 there is exactly one value of x ~ 0 such that y =
x".
(Later on we shall see how this property may be deduced from the theory of continuous functions.) If y ~ 0, the value of x ~ 0 which satisfies the equation y = x" is called the nth root of y and is denoted by x = yl/n.
vy
= yll2. Note that, with this conWhen n = 2, we also use the notation vention, it is always true that O. If Y > 0, there are, of course, two numbers whose square is y. The positive one is and the negative one is  Vy. The notation ± Jv means 'Vy or  Vy'. If r = mln is a positive rational number. and y ~ 0, we define
J» ~
yr
=
J»
(ym)lln.
If r is a negative rational, then r is a positive rational and hence yr is defined. If y > 0 we can therefore define yr by +
yr
=
_1_ yr'
We also write yO = 1. With these conventions it follows that, if y > 0, then yr is defined for all rational numbers r. (The definition of yX when x is an irrational real number must wait until a later chapter.)
1.10
Quadratic equations
If y > 0, the equation x2 = y has two solutions. We denote the positive solution by Vy. The negative solution is therefore Vy. We note again that
Real numbers
7
there is no ambiguity about these symbols and that ± Vy'. The general quadratic equation has the form ax"
+ bx + c
Vy simply means
'Vy or
= 0
where a =1= O. Multiply through by 4a. We obtain 4a2x2
+ b)2 b2 + 4ac + b)2 = b2 4ac.
(2ax (2ax 'I
~
+ 4abx + 4ac =
+b x
=
± V(b2 4ac)
b
2a
+ bx + c = 0 are therefore 4ac) b + V(bl 4ac) ~ = _....:....>_'
The roots of the equation ax2
I
a =
 b  V(b2
+
and
2a
II
jl
0
= ± V(b2  4ac)
(
l
=
It follows that the quadratic equation has no real solutions if b2  4ac < 0, one real solution if b2  4ac = 0 and two real solutions if iJ2. 4ac > O. If b2 4ac ~ 0, 2ax
I
0
2a
It is a simple matter to check that, for all values of x, ax2
+ bx + c
= a(x  a)(x  ~).
With the help of this formula, we can sketch the graph of the equation y = ax" + bx + c.
y = ax' + bx
+c
a>O
1.11 Example A nice application of the work on quadratic equations described above is the proof of the important CauchySchwarz inequality. This asserts that, if ai, a2, ... ,an and b 1, b2, ... .b'; are any real numbers, then. (a1b1 +a2b2
+ ... + anbn)2
:s:;;;;
(ar
+ a~ + ... + a~)(br + b~ + .. , + b~).
Real numbers
8
Proof For any x,
o ~ (alx + bl)2 + (a2X + b2)2 + =
Since y
+ (anx + bn)2
(ar + ... +a~)x2 + 2(alb, +
+anbn)x + (br + ... + b~)
= Ax2 + 2Bx + C. = Ax2 + 2Bx + C ~ 0 for all values of x,
it follows that the equation
AX2 + 2Bx + C = 0 cannot have two (distinct) roots. Hence (2B)2  4AC ~ 0 B2 ~ AC
i.e.
which is what we had to prove.
Exercise
1.12
(1) Suppose that n is an even natural number. Prove that the equation xn = y has no solutions if y < 0, one solution if y = 0 and two solutions if y > O. Suppose that n is an odd natural number. Prove that the equation xn = y always has one and only one solution. Draw graphs of y = x2 and y = x3 to illustrate these results. (2) Simplify the following expressions: (i) 82/3 (3) If y ing:
(ii) 274/3
> 0, Z > 0 and rand
(iii) 326/5•
s are any rational numbers, prove the follow(iii) (yz
Y = yr zr .
(4) Suppose that a > 0 and that 0: and {3 are the roots of the quadratic equation ax? + bx + c = 0 (in which b?  4ac > 0). Prove that y = ax? + bx + c is negative when 0: < x < {3 and positive when x < 0: or x > {3. Show also (without the use of calculus) that y = ax2 + bx + c achieves a minimum value of c  b'214awhen x =  sn«. (5) Let a I, az, ... ,an be positive real numbers. Their arithmetic mean An and harmonic mean Hn are defined by
An
=
al+a2+
... +an n
1(11
1]
H;; I = .  +  + ... +  . n al
a2
an
Deduce from the CauchySchwarz inequality that Hn ~ An. (6) Let ai, a'2,' .. ,an and bl, b'2," . .b ; be any real numbers. Prove Minkowski's inequality, i.e.
1)
L~, )'!'" .; It (a. + bk
]"'+
a1
(
n ]"2 Lb~
k ::1
Real numbers
9
For the case n = 2 (or n = 3) this inequality amounts to the assertion that the length of one side of a triangle is less than or equal to the sum of the lengths of the other two sides. Explain this.
1.13
Irrational numbers
In § 1.2 we mentioned the existence of irrational real numbers. That such numbers exist is by no means obvious. For example, one may imagine the process of marking all the rational numbers on a straight line. First one would mark the integers. Then one would move on to the multiples of ~ and then to the multiples of! and so on. Assuming that this program could ever be completed, one might very well be forgiven for supposing that there would be no room left for any more points on the line.
r'.!!I1!1!N1.!."HM~ ••
"'!!!!lUI!!~!I!!I!!II1!'''!~!!Itn!II!1''"'I!!I!1I!I!''!!ltpi''"l!"M""!'''I!!!1!~I''lIIttrlMt!!'''''!!!1

1
t
0
t 1 1
I
1 t 1
2
But our assumption about the existence of nth roots renders this view untenable. This assumption requires us to accept the existence of a positive real number x (namely y2) which satisfies x2 = 2. If x were a rational number it would be expressible in the form
x=
m n
where m and n are natural numbers with no common divisor (other than 1). It follows that
and so m? is even. This implies that m is even. (If m were odd, we should have m = 2k + 1. But then m2 = 4k2 + 4k + 1 which is odd.) We may therefore write m = 2k. Hence 4k2 = 2n2 n2 = 2k2•
Thus n is even. We have therefore shown that both m and n are divisible by 2. This is a contradiction and it follows that x cannot be rational, i.e. y2 must be an irrational real number. Of course, J: is not the only irrational number and the ability to extract nth roots allows us to construct many others. But it should not be supposed that all irrational numbers can be obtained in this way. It is not even true that every irrational number is a root of an equation of the form
10
Real numbers .ao + aJx
+ a2x2 + ... + a.x"
= 0
in which the coefficients ao, aI, ... .a.; are rational numbers. Real numbers which are not the roots of such an equation are called transcendental, notable examples being e and 1T. This fascinating topic, however, lies outside the scope of this book.
1.14
Modulus
Suppose that x is a real number. Its modulus (or absolute value) [x I is defined by
(x~o) (x
< 0).
y = Ixl
x
o
Thus 131= 3,161 = 6 and 101 = O. Obviously [x] ~ It is sometimes useful to note that Ix I = .jx2. 1.15
0 for all values ofx.
Theorem For any real number x, Ixl
~ x ~ [x].
,
Proof Either x ~ 0 or x < O. In the first case, Ixl the second case, Ixl = x < 0 < [x],
~ 0 ~ x = [x], In
I
i /
1.16
Theorem
I
For any real numbers a and b
lab I = lal.lb I Proof The most elegant proof is the following. labl
=
.j«ab)2)
=
.j(a2b2)
=
.j(a2) • .j(b2)
=
lal.lbI.
The next theorem is of great importance and is usually called the triangle
inequality, This nomenclature may be justified by observing that the theorem is the special case of Minkowski's inequality (exercise 1.12(6)) with n = 1. We give a separate proof.
1.17
Theorem (triangle inequality) la + bl ~ lal
+
Proof We have
Ibl.
For any real numbers a and b,
I
Real numbers la+bl2
=
11
(a+b)2
It follows (from example
= a2+2ab+b2 = lal2 + 2ab + Ibl2 ~ lal2 + 21abl + Ibl2
(theorem l.IS)
=
lal2 + 2lal.lbl + Ibl2
=
(Ial + Ibl)2.
(theorem
1.16)
1.6) that
la + bl ~ lal + Ibl. 1.18
Theorem For any real numbers c and d,
Ic dl
~ IclIdl.
Proof Take a a + b = c and so lei 1.19
=
la + bl ~ lal + Ibl
=
d in the triangle inequality. Then
Idl + Ie dl.
Example (i) 2 = (ii) 7 =
1.20
= d and b = c 
11 + 31 ~ 111 + 131 = 1 + 3 = 4. 16  (1) I ~ 161 1 I I = 6  I = 5.
Exercise (1) Prove that lal < b if and only if =b O} has no maximum, nor does it have any upper bounds. The set {x :x > O} has no minimum (why not?). Its largest lower bound is O.
2.9
Intervals
An interval I is a set of real numbers with the property that, if x E I and y E I and x ~ z ~ y, then z E I, i.e. if two numbers belong to I, then so does every number between them. In describing intervals we use the following notation:
(a,b) [a,b]
= =
{x:a<x 0,
~x ~ ~B for any xES. Hence T is bounded above by ~B. By the Continuum Property, Thas a smallest upper bound (or supremum) C. We have to prove that C = ~B. Since ~B is an upper bound for T and C is the smallest upper bound for T, C~ ~B. Now repeat the argument with the roles of Sand T reversed. We know that C is the smallest number such that, for any yET,
y~c. Since ~ > 0, it follows that ~ly ~
~1
C
for any yET. But S= {~1Y :yET}. Hence ~1 C is an upper bound for S. But B is the smallest upper bound for S. Thus B ~ ~IC ~B ~ C. We have shown that C ~ ~Band also that ~B~ C. Hence ~B = C.
2.13
Exercise (1) Suppose that S is bounded above and that So C S. Prove that sup So ~ sup S, (2) Suppose that S is bounded above and that ~ is any real number. Prove that sup (x xES
+ ~)
sup x xES
+~.
Continuum property
19
(3) Suppose that S is bounded above. Prove that inf (x)
= 
xES
sup x. xES
Hence obtain analogies of theorem 2.12 and exercises 2.13(1 and 2) in which inf replaces sup. (4) The distance d(t S) between a real number ~ and a nonempty set S of real numbers is defined by d(~,S)
= inf
I~xl
xES
(see exercise 1.20(3». Find the distance between the real number ~ = 3 and the sets (i) S = {O, 1, 2} (ii) S = (0,1) (iii) S = [1,2] (iv) S = (2, 3). (5) (i) If ~ E S, prove that d(t S) = O. Give an example of a real number ~ and a nonempty set S for which d(t S) = 0 but ~ f/. S. (ii) If S is bounded above and ~ = sup S, prove that d(t S) = o. Deduce that the same is true if S is bounded below and ~ = inf S. t (iii) If I is a closed interval, prove that d(~, I) = 0 implies that ~ E I. If I is an open interval (other than IR or f/J), show that a ~ q:. I can always be found for which d(t I) = O. * (6) Suppose that every element of an interval I belongs to one or the other of two nonempty subsets Sand T. Show that one of the two sets S or T contains an element at zero distance from the other. (Do not assume that S and Tare necessarily intervals. The set S, for example, could consist of all rational numbers in I and the set T of all irrational numbers in I.) [Hint: Suppose that s < t, where s E Sand t E T. Let To = {x : x E T and x> s}. Show that To =1= f/J and is bounded below. Write b = inf To and consider the two cases bET and b q:. T.]
3
NATURAL
3.1
Introduction
NUMBERS
In this chapter we explore some implications of the Continuum Property for sets 01 natural numbers. As always, arithmetical properties are taken for granted. But where a proof hinges on such a property, the property will be explicitly stated.
3.2
Archirnedean property The Archimedean property is the assertion of the following theorem.
3.3
Theorem The set N of natural numbers is unbounded above.
Proof For this proof we need to assume that, if n is a natural number, then so is n + 1. Suppose that the theorem is false. Then N is bounded above. By the Continuum Property it therefore has a smallest upper bound B. Since B is the smallest upper bound for N, B1 is not an upper bound for N. Thus, for some nEN,
n >Bl n + 1 > B. But then n + 1 is an element of the set N which is greater than B. But B is an upper bound of the set N. This is a contradiction. Hence N is unbounded above.
3.4
Example Prove that the set S = {n1: n EN}
is bounded below with largest lower bound
o.
Proof Since the natural numbers are all positive,
10
+1++1
t+t+t~ s
20
21
Natural numbers
for all n EN. Hence the set S is bounded below by O. We have to prove that 0 is the largest lower bound. Suppose that h > 0 is a lower bound for S. Then, for each n E N,
Hence, for each n EN, n Xn (n = 1,2, ... ). It is strictly decreasing ii xc ., < Xn (n = 1,2, ... ). A sequence which is either increasing or else decreasing is called monotone.
4.16
Example
(i) (ii) (iii) (iv) decreasing.
4.17
The The The The
sequence (2n) is strictly increasing. sequences tn 1) and ( n) are both strictly decreasing. sequence (1) is both increasing and decreasing. sequence «( 1)") is not monotone. It is neither increasing nor
Theorem
(i) If (xn) is increasing and bounded above, then it converges to its smallest upper bound. (ii) If (x n> is decreasing and bounded below, then it converges to its largest lower bound.
Proof If (xn) is decreasing and bounded below, then ( xn) is increasing and bounded above. Hence it is only necessary to prove (i). Suppose that (xn) increases and is bounded above with smallest upper bound B. We must show that Xn + B as n + Let E > 0 be given. Since B  E is not an upper bound, there exists at least one term XN of the sequence such that 00.
xN>BE. But (xn) increases. Hence, for any n
Xn ~xN>BE.
> N, x., ~
XN.
Thus, for any n
> N,
35
Convergent sequences Also B is an upper bound for the sequence. Hence Thus, for any n > N,
i.e.
BE<Xn
~B.
BE<Xn
0, we have found an N such that, for any n Hence Xn ~ Bas n ~ 00.
> N,
IXn  BI
< E.
BriII~~~ '1I I I
I I I I I I
I I
I I
I
N
4.18 Examples (i) The sequence «(n  1)/ n>increases and we know from exercise 3.6(1) that it is bounded above with smallest upper bound 1. By theorem 4.17 it follows that nl 
+
i as n
+
00.
n
(ii) If 0 < x < I, the sequence (xn> decreases and is bounded below with largest lower bound 0 (exercise 3.6(2)). Thus
4.19 Example sequence
The examples above are nothing new. But consider the
«(1 + nl)n>. We first show that this sequence increases. In the inequality for the geometric and arithmetic means, take al = a2 = ... = anl = 1+ (n _1)1 and an = I. Then
36
Convergent sequences 1 ) and so we obtain the contradiction I = 1 and I =  1.
n
+
00.
°
We say that a sequence (xn) diverges to + and write Xn for any H> 0, we can find an N such that, for any n > N, 00
+
+
00
as n
+
00
if,
Convergent sequences
39
Xn >H. Similarly, a sequence (xn) diverges to  00 and x n + H> 0, we can find an N such that, for any n > N,

00
as n
+
00
if, for any
xn which does not satisfy xn + + 00 as n + 00 or 00 as n + 00 is said to oscillate. An example is the sequence «( l)").
+ 
4.28 n + 00.
Example Let
Q
Proof Let H
>
be a positive rational number. Prove that n"
°
+
+
00
as
be given. We have to find an N such that, for any
n>N, nCt.>H i.e.
n
> HI/Ct..
We simply choose N = H"Ct..Then, for any n
> N, n > HI/a
and hence nCt.> H.
A common error is to suppose that proposition 4.8 (about combining con00. For example, it is tempting to suppose that, if x.; + 00 as n + 00 and j/., + 00 as n + 00, then xn  Yn + 00  00 = as n + 00. But 00 is not a real number and cannot be treated as such. Indeed, one should immediately be warned that something is wrong by the appearance of the totally meaningless expression 00  00. To press the point home, we consider the example x., = n2 (n = 1,2, ... ) and y., = n (n = 1,2, ... ). Then x., + 00 as n + 00 and Yn + 00 as n + 00, cut Xn  Yn + 00 as n +00. Or, again, take x., = (n + 1) (n = 1,2, ... ) and y., =n (n = 1,2, ... ). This time xn  Yn + 1 as n + 00.
vergent sequences) applies to sequences which diverge to + 00 or 
°
f'} J
I! ,I
40
Convergent sequences
4.29
Exercise (1) Suppose that Xn ~ I as n ~ (i) IXn II ~ 0 as n ~ 00 (ii) IXnl ~ III as n ~ 00. (2) Suppose that xn ~ I as n ~ that, for any n > N, Xn
00.
Prove that
00.
If I> 0, prove that there exists an N such
> !/.
(3) Prove that (ii) 00 as n ~ 00 (i) 2n ~ + 00 as n ~ 00 (iii) «(It/!) oscillates. (4) Let xn > 0 (n = I, 2, ... ). Prove that x., ~ 0 as n ~ 00 if and only if l/xn ~ 00 a~~n ~ 00. (5) Suppose that (xn) increases and is unbounded above. Prove that Xn ~ + 00 as n ~ 00. If (xn) decreases and is unbounded below, prove that Xn ~  00 as n ~ 00. * (6) Let S be a nonempty set of real numbers and suppose that d(~, S) = 0 (see exercise 2.l3(4)). Show that, for each n EN, we can find an x; E S such that I~ xnl < l/n. Deduce that Xn ~ ~ as n ~ 00. If S is bounded above, show that a sequence of points of S can be found which converges to its supremum. If S is unbounded above, show that a sequence of points of S can be found which diverges to + 00.
vn ~ 
5
SUBSEQUENCES
5.1
Subsequences
Suppose that (xn) is a sequence and that ( n; ) is a strictly increasing sequence of natural numbers. Then the sequence (xn ) r
is called a subsequence of (xn>. The first few terms of some subsequences of
X2, X4, X6,
(X3r+s>
= = =
(X2r>
=
X2,X4,XS,X16,···
(Xr+1> (X2r>
(xn
> are listed below.
X2,X3,X4,··· .••
xS,XU,X14,X17,'"
Roughly speaking, if we think of a sequence as a list of its terms, then we obtain a subsequence by crossing out some of the terms. For example, (X2n> may be obtained from (x n > by crossing out terms as below.
J
The insistence that t.n; > is strictly increasing is of some importance. It means, for example, that the sequence whose first few terms are
is not a subsequence of (xn>. The fact that ( n; > is strictly increasing has the consequence that n; ~ r
(r = 1,2, ... ).
This is easily proved by induction. 5.2 of (xn
Theorem Suppose that
Xn +
I as n +
00
and that (x,) is a subsequence
>. Then Xn
r
+ I
as r
+
00.
Proof Let E > 0 be given. Since x., + I as n that, for any n > N,
+
00,
there exists an N such
41
42
Subsequences
Ix., II < €. Let R = N. Then, for any r
> R,
Ixnll R,
nr;;:' r > R = N and so n;
> 0, we can
> N.
Thus
find an R (namely R
= N)
such
Ixnrll 0, then xlln 1 as We prove this again by a different method. As in example 4.4 we need only consider the case x ;;:.1 . If x;;:. 1, then xlln ;;:. 1 (n = 1, 2, ... ) and hence the sequence < x'!") is bounded below by 1. Moreover
5.4 n
jo
jo
00.
xlln/xl/(n+l)
= x(n+ln)/n(n+l)
= x1ln(n+1);;:'1
(n = 1,2, ... )
and therefore (Xlln) decreases. From theorem 4.17 it follows that t.x'!") converges to a limit 1 and 1 ;;:. 1. What is the value of /? By theorem 5.2, we know that all subsequences of (Xl In) must also tend to /. Hence Xl/2l1jo/asnjooo. Using proposition 4.8, it follows that xlln
=
x'l?": x'l?" jo/.1
=
/2 as n
jo
00.
But a sequence can only have one limit. Thus
/ = 12
°
and so / = or 1 = 1. But I;;:' 1 and hence 1 = 1. We have shown that xllnjolasnjooo.
Note: A warning is appropriate here. Do not use this method of working out the value of a limit until you have shown that the sequence converges. For example, if you thoughtlessly tried to evaluate the limit of the sequence «(1)n) by looking at its subsequences, you would obtain / = + 1 =  1.
Subsequences 5.5 Example inductively by
Let a
Xn+! = !exn
Obviously Xn
43
> 0 and let Xl>
+
... ). But further
+a
_2 xn
2 xn+!xn
(n = 1,2, ... ).
ax;;!)
> 0 (n = 1,2,
O. Define the rest of the sequence (xn >
We know in advance that this quadratic equation in x.; has a real solution. Thus 'b2  4ac;> 0', i.e.
> 0, it follows ;> va (n =
that
Since xn+! xn+l
We wish to prove that Xn  Xn+l
(xn)
1,2, ... ). decreases and so we consider
= Xn  HXn
+ ax;;l)
;> 0 (n
2,3, ... ).
=
Hence the sequence (xn) decreases and is bounded below by we omit its first term). From theorem 4.1 7 it follows that X n +
1 as n
+
Va (provided
00
va.
where I;> What is the value of I? By theorem 5.2 we also have Xn+l + 1 as n xn+!
= HXn
Now 1:::/=0 (because I;> xn+l
= !exn
+
+
But
00.
ax;;!).
va). Hence, by proposition + ax;;!)
+
~(l
+ aiI)
as n
+
4.8, 00.
Since a sequence can have at most one limit, I = t(l i.e.
+ arl)
12 = a.
va
Thus 1 = or 1 = therefore shown that Xn
+
va as n
va. But we know that I;> va. Hence 1 = va. We have +
00.
Subsequences
44
5.6 Example We attack the previous example by a different method. Again let a> 0 but this time let Xl be unspecified for the moment. Define the rest of the sequence inductively by Now
xn+l = HXn +ax;;l)
(n = 1,2, ... ).
Xn+l  va = !exn + ax;;l)  va
1
1
2Xn (2Xn_1
=
1 2 (2 ) Xn2 )4 (Xn2
 va)8
1 1 1 . / 2n 2x.; (2X _ )2' .. (2XI )2n1 (XI  va) . n 1
If we assume that XI ~ Ja, then it follows, as in example 5.5, that Xn ~ va = 1, 2, ... ). Hence
(n
IXn+1  va I is a bounded sequence and let L denote the set of all real numbers which 'are the limit of some subsequence of ( x n ). We know from the BalzanoWeierstrass theorem that L is not empty (i.e. L =1=0). Of course, if (xn) converges, L will just consist of a single point (theorem 5.2).
5.13t
Proposition Let < xn) be a bounded sequence and let L be the set of all real numbers which are the limit of some subsequence of < xn). Then L has a maximum and a minimum. That L is a bounded set follows from theorem 4.23. Hence L certainly has a supremum and infimum. Proposition 5.13 asserts that these actually belong to the set L. The proof is relegated to the appendix. Readers who prefer to omit
48
Subsequences
the proof altogether will encounter it again at a later stage since it is a special case of a general theorem concerning 'closed sets'. _ If (xn) is a bounded sequence, let us denote the maximum of L by I and the minimum by 1. Then 7 is the largest number to which a subsequence of (xn) converges and { is the smallest number to which a subsequence of (xn) converges. We calli the limit superior of (xn ) and write
I =
lim sup
Xn•
n+oo
Similarly, {is the limit inferior of (xn)
and we write
{ = lim inf Xn. n+oo
(Occasionally you will see such expressions as lim sup n+oo
means that
(xn)
x., = +
is unbounded above. Similarly lim inf
xn = 
00.
00
This simply means that
n+OO
(xn)
5.14 t
is unbounded below.)
Examples
(i) The sequence (1In) converges with limit O. By theorem 5.2, all its subsequences converge to O. Hence L contains the single point 0 and 1 1 1 lim sup  = lim inf  = lim  = O. n+oo n n + n n>oo n 00
(ii) The sequence «(  l )") is bounded. The set L consists of two points and 1 (i.e. L = {+ 1, 1}). We have limsup(I)n
= l;liminf(I)n
+1
=1.
n+oo
(iii) Let ( r n) be the sequence whose first few terms are t l, i,t t 1, t, ~,t 0 < r n < 1 (n = 1,2, ... ). From theorem 4.23 it follows that L is a subset of [0, 1] (i.e. L C [0, 1D. But the subsequence !, l. con~erges to 0. Thus L = 0. Also the subsequence t i, t t ...converges to 1. Thus I = 1. We have shown that
t t ...Obviously
lim sup r n = 1;
t ~,...
lim inf rn =
n>oo
5.15 t
°
Exercise
t (1) Calculate (i) lim sup {(l)n(l
+
n1)}
(ii) lim inf {(l)n(l n+oo
+ n1)}
Subsequences
49
and show that these are not the same as the numbes
+ n1)
sup (I)n(1
and
n~1
t (2) Let
inf ( l)n(1
+ n1).
n>1
< rn) be the sequence whose first few terms are
i,i, ... Show that
t t i,t a, t t ~, !,
every point in the interval [0, 1] is the limit of a subsequence of < rn). [Hint: see exercise 3.6(4).] t (3) Suppose that (xn) is a bounded sequence and that, for any N, we can find an n > N such that xn ~ b. Show that (xn) has a subsequence which converges to a limit I ~ b. t (4) Let (xn) be a bounded sequence with limit superior / and limit inferior i. Show that, given any E > 0, we can find an N such that, for any n > N, Xn is a Cauchy sequence if, given any E > 0, we can find an N such that, for any m > N and any n > N,
I xm xn 1< E. Very roughly, the terms of a Cauchy sequence get 'closer and closer' together. 5.17
Proposition
Any convergent sequence is a Cauchy sequence.
5.18
Proposition
Any Cauchy sequence is bounded.
The proofs are easy. The next theorem is what makes Cauchy sequences important. 5.19
Theorem Every Cauchy sequence converges.
Proof Let (xn> be a Cauchy sequence. By proposition 5.l8, (xn> is bounded. Hence, by the BolzanoWeierstrass theorem, it has a convergent subsequence (xn r >. Suppose that Xn r * I as r * oc;. We shall show that x n * I as n * Let E > O. Then h > O. Hence there exists an R such that, for any r > R, 00.
IXn
r
ll is a Cauchy sequence, there also exists an N such that, for any m and any n >N,
>N (2)
Now suppose that n > N and choose r so large that n; > Nand r > R. Then (1) is satisfied and also (2) is satisfied with m = n.. Thus, for any n > N,
I Xn

x.; r + Xn r II
~ IXn

Xn
< !E + tt
r
=
I + IXn r II
(triangle inequality)
E.
Given any E > 0, we have found an N such that, for any n Thus x., * I as n * 00.
> N, IXn II < E.
Subsequences 5.20
51
Example A sequence (xn> is defined by
=
Xn+2
Prove that (xn)
i{xn+l
=
(n
+xn)
1,2,3,
Xl
= a, X2 = band
... ).
converges.
Proof We have xn+2 
Thus
= i{xn+l
Xn+l
+ xn)

=
Xn+l
i{xn

Xn+l)'
.11 =
Ixn+2xn+11
Hence, if n
=
2"IXnXn+11
22IxnXn11
> m,
I x., 
Xm
I

Xnl
+ Xnl
~ I x., 
Xnl
I + I Xnl
=
I Xn

Xn_2 
xn21
...
+ Xm+l

i
Xm
+ ... + I Xm+l

Xm
1
~ (2~2+ 2~3+ ... + 2LI) Ib a I ~ 2!' (1+ 1
E
+ ;, + ... +
1  (1)nm .
= 2m1
Let
±
1
_!
Ib a
2"!m.) I b  a i 1
1
~2m2Ib
+
a I.
> 0 be given. Choose N so large that 1
2 N2
1
b a
Then, for any n
I Xn
1
< E.
> N and
xm
I~
any m
1 2Ni1b
> N.
a I < E.
Thus (xn > is a Cauchy sequence. Therefore, by theorem 5.19, it converges. (An alternative method would be to show that one of the two sequences (X2nl> and (X2n > is increasing and the other decreasing and then to show that they have the same limit.)
5.21
Exercise (1) Suppose that 0 < a < 1 and that (xn) is a sequence which satisfies I xn+l  x.; I ~ an (n = 1,2, ... ). Prove that (xn > is a Cauchy sequence and hence converges.
52
Subsequences Give an example of a sequence (Yn > such that Yn ~ + but I Yn+l  Yn I ~ 0 as n ~ [Hint: see exercise 4.20]. (2) A sequence (xn> satisfies 0 is a Cauchy sequence and hence converges. (3) For the sequence of example 5.20, prove that
Deduce that Ixn+l /1 = t IXn II, where / = HX2 + iXl)' What conclusion may be drawn about the convergence of the sequence (xn)? Tackle exercise 5.21(2) by a similar method. (4) Let [a, b] be a compact interval (see §2.9). Prove that every sequence of points of [a, b] contains a subsequence which converges to a point of [a, b]. t (5) Let I be an interval which has the property that every sequence of points of I contains a subsequence which converges to a point of I, Prove that I is compact. [Hint: First prove that I is bounded by assuming otherwise and appealing to exercise 4.29(6). Then prove that sup I and inf I are both elements of I with another appeal to exercise 4.29(6).] t (6) Given a set S of real numbers, let St = {x:xESandx*~}. We say that ~ is a cluster point (or point of accumulation or 'limit point') of S if ~ is at zero distance from St' (Note that ~ need not be an element of S). A form of the BolzanoWeierstrass tneorem asserts that every bounded set with an infinite number of elements has at least one cluster point. Prove this.
6
SERIES
6.1
Definitions Given a sequence of real numbers the sequence <SN>defined by N
SN =
L n=1
an = al
+ a2 + ... + aN
is called the sequence of partial sums of the series
If SN + S as N
+
00,
the series is said to converge
to the sum s. We write
s = fan. n=1
It is important to remember that this formula can only make sense when the series converges.
Example
6.2
Consider the series
00
L
x".
neO
If x
=1=
1, the partial sums satisfy N
SN
= L
x"
=
I+x
+ x2 + ... + xN
n=O
I __ XN+1
= If [x] < I,
XN+1
+
0 as N
Ix +
00
and hence
53
54
Series 1
sN+ 
1 x
as N +
00.
It follows that the series 1 =.1 x
I xn 00
n=O
L;;'
=0
(lx]
x" converges if [x I < 1 and we may write
< 1).
If lx] ;;;.1, the series diverges.
6.3
Example
I
n=!
The partial sums of the series 1
00
n(n
+ J)
are given by
J,n(n~l) 1(;n~l) (I~)+ (~t) + (t±) + +(h I)
'N=
N~
1 1  + 1 as N + N+ 1
00.
Hence the series converges and we may write
I
1
00
n=!
6.4
n(n
+ 1)
= l.
Series of positive terms
Series whose terms are all positive (or nonnegative) are particularly easy to deal with. This is because the sequence of partial sums of such a series is increasing. Thus, if we wish to show that a series of positive terms converges, we only need to show that its sequence of partial sums is bounded above. If the sequence of partial sums is unbounded above, then the series diverges to + (see exercise 4.29(5)). 00
6.5
Theorem The series 1
00
In
n=!
diverges to
+
00.
55
Series Proof Since the partial sums of this series increase, we only need to show that they are unbounded above. But 1
1
1
= 1 +"2+3+"'+
S2N
2N
~ I + i+ (~+ ~1+ (~+ ~+~+~) +... +eN_II+ I + .. + ;N) ;, l+t+ (~+~) + (~+~+~+~) +.+(> ... +2~) 1
2
2N1
4
= 1 + 2 + 4 + 8 + . . . + 2N = 1 + IN 2
•
Thus the partial sums are unbounded above and the theorem follows.
6.6
Theorem Let a be a rational number such that a> 1. Then the series 00
1
Lei
n=ln
converges.
Proof Since the partial sums are increasing, we need only show that they are bounded above. For N > 1, 1
SN ~S2N_I
1
= 1 + 2a + 3a + ... + (2N
1 l)a
~ I + (2~+ 3~1+ (4~+5~+6~+;. )+. + (2(N~1,"+ ... + 1)0) (2N
~ 1+
Ua + 21a) + (4~
+
(2(N~l)a
2
= 1
=
=
~
+ 4~ + 4~ + 4~) + ...
+ ... + 2(N~l)a) 2NI
4
+ 20, + 40, + ... + 2 (N 1)0,
1 1++ 20,1 1 (1/2aIyN t (l/2al)
(1)2 20,1,
. (
1 ~N I + . . . +20,1 1
~:a l I (l/2  )
56
Series
provided that 1/2IY. 1 < 1. But this follows from the assumption that Hence (SN) is bounded above and the theorem follows.
6.7
a>
1.
Elementary properties of series
6.8 Theorem Suppose that the series ~;;' = 1 an and ~;;' = 1 b.; converge to a and {3 respectively. Then, if A and J.1. are any real numbers, the series ~;;' = 1 (Aan + J.1.bn) converges to Aa + J.1.{3. Proof
We have
N
L (Aan
n=1
+ J.J.f3n) = A +
Aa
N
N
.. =1
n=1
L an + J.1. L s;
+ J.J.f3as N +
00
by proposition 4.8. 6.9
Suppose that the series ~;;' = 1 an converges. Then
Theorem an
+
0 as n
+
00.
Let the sum of the series be s. Then
Proof
N
SN
= L
an
+
s as n
+
00.
n=1
Also
SNl+S
asn
+00.
But then
6.10
,. I.
Examples
The series
f ( l)n
n=1
diverges. This can be deduced from theorem 6.9 by observing that its terms do not tend to zero. Note that the converse of theorem 6.9 is false. Just because the terms of a series tend to zero it does not follow that the series converges. Theorem 6.5 provides an example. The terms of the series 00
1
Ln=1 n
Series
57
tend to zero, but the series diverges. One might say that the terms of the series do not tend to zero 'fast enough' to make the series converge.
6.11
Proposition Suppose that the series L;;'= 1 an converges. Then, for each natural number N, the series L;;'=N an converges and 00
L
n=N
an ~ 0 as N ~
00.
This result is often referred to by saying that the 'tail' of a convergent series tends to zero. The proof is easy.
6.12
Series and Cauchy sequences
In the series we have studied so far, we have either been blessed with a nice formula for the partial sums (examples 6.2 and 6.3) or else the partial sums increased and so we only had to consider whether or not the partial sums were bounded above (theorems 6.5 and 6.6). What should we do in the absence of such favourable conditions? We ask the question: is the sequence of partial sums a Cauchy sequence? This question has some fruitful answers as we shall see below. 6.13 Theorem Suppose that (an> is a decreasing sequence of positive numbers such that an ~ 0 as n ~ Then the series 00.
00
'\' L..
n=1
(I)nlan
=
ala2
+ a3a4
+ ..•
converges.
Proof We show that the sequence (sn> of partial sums of the series is a Cauchy sequence. From theorem 5.19, it then follows that the series converges. The proof depends on the fact that, for each n > m, we have the inequality
This follows easily from the fact that ak (ak> decreases. Let e > 0 be given. Since an ~ 0 as n ~ n >N,an < e. But for any n >m >N, Isnsml
=
l(ala2+a3
~ am+1
<e
ak + 1 is
00
always nonnegative because
we can find an N such that, for any
... an)(ala2+
(because m
> N).
Thus (sn> is a Cauchy sequence and the theorem follows.
... am)1
58
Series
6.14 Example We saw in theorem 6.5 that the series L;=I I/n diverges. But it follows from theorem 6.13 that the series 00
(_
1)n  1
L 
n=1
1.
I
I
2
3
4
= 1++
n
I
 ... 5
converges. Of course, theorem 6.13 yields no clue as to what the sum is. (It is, in fact, 10&12.)
Theorem 6.13 is sometimes useful, but its hypotheses are rather restrictive. A more useful theorem is the following. 6.15 Theorem (comparison test) Let L; = 1 b.; be a convergent series of positive real numbers. If (n = 1,2, ... )
lanl ~ b ;
then the series L;=I an converges. Proof Let E > O. Since L;=I b ; converges, its tail tends to zero (proposition 6.11). Hence we can find an N such that, for any n > N, 00
L
k=n+1
bk
< E.
Let the sequence of partial sums of the series L; n>m>N,
~
lam+11
~ bm+1
= 1 an
+ lam+21 + + lanl + bm+2 + + b.;
be (sn). Then, if
(triangle inequality)
Thus (sn) is a Cauchy sequence and the theorem follows. Note that the hypothesis lanl ~ b., (n = 1,2, ... ) can be replaced by lanl ~ Hb ; (n = N, N + I, ... ). (Why?) Note further that the hypothesis of the comparison test is not n
L ak hool
n
~
L bk· hool
This condition is satisfied by ak l:;=1 (I)n diverges.
= (I)k
and bk = l/k2,
but we know that
59
Series Prove that, if
IX
is any positive rational number and [x] < 1,
converges.
Proof From exercise 4.20(5) we know that nOl+2xn ~ 0 as n ~ Hence the sequence (nOl+2xn) is bounded (theorem 4.25). It follows that, for 00.
some H,
InOlxnl ~ H/n2
(n = 1, 2, ... ).
Since the series L;;'=I 1/n2 converges (theorem 6.6), the result follows from the comparison test.
The next two propositions are sometimes helpful. We indicate only very briefly how they are proved. 6.17
Proposition lim n +
00
\aan
n 1 +
(ratio test) Let \
L;;'=I
an be a series which satisfies
= I.
If I> 1, the series diverges and, if 1< 1, the series converges. If 1< 1, we may take large value of N,
€
> 0 so small that I + < 1. Then, €
N 2 lanl =  an Ilanil ... la + laN+II aN+I anI an2 j
I
for a sufficiently
< (I + e)nNI
laN+d.
The series L;;'= I an then converges by comparison with the geometric series L;;'= I (I + e)", If I > 1, a similar argument shows that the terms of L;;'= I an do not tend to zero and so the series diverges. 6.18
Proposition
(nth root test)
Let
L;;'=I
an be a series which satisfies
lim sup lanilln = I.
n+oo
If I> 1, the series diverges and, if 1 < 1, the series converges. If 1 < 1, we may take e > 0 so small that 1 + e large value of N,
lanl«/+€r
(n>N)
< 1. Then,
for a sufficiently
(exercise 5.15(4))
and the convergence of L;;'= I an follows from the comparison test. If I> 1, e is
60
Series
chosen so that I lankl
E
> 1. Then,
> (/
Etk ~
k;;'=
and so the terms of
1
00
for some subsequence as k ~
(ank),
00
an do not tend to zero.
Note that if the expressions of propositions 6.17 and 6.18 diverge to + (instead of converging to I), then the series diverge. If 1 = 1 however the results yield no information about the convergence or divergence of the series at all. 00
6.19
Example 00
Prove that the series
x"
Ln!
n=o
converges for all values of x. Proof
We could use the comparison test. Alternatively, if x =I=0,
x+ /x n 1 +
(n
n
1)!

n!
[x] = 
n
+1
~ 0 as n ~
. 00
and hence the series converges by the ratio test.
6.20
Absolute and conditional convergence
A series ~;;'=l an is said to converge absolutely if the series k;;'~l lanl converges. A series which converges but does not converge absolutely is said to be conditionally convergent. 6.21
Theorem Proof
6.22
Every absolutely convergent series is convergent.
Simply take b; = lanl in the comparison test.
Examples (_l)nl (i) Let an =
n
(n = 1, 2, ... ). Then 111 1++ 2 3
00
L lanl n=!
00
=
L n=l
n
4
...
61
Series
. The first of these series converges (example 6.14). The second diverges (theorem 6.5). We conclude that the series (_1)nl
L '00
n
n=1
is conditionally convergent. (ii) On the other hand the series 00
00
L
an
n=1
(_
= L n=1
1)n  1 n
2
is absolutely convergent because
which converges.
It should be noted that the comparison test, the ratio test and the nth root test all demonstrate absolute convergence. The only criterion we have given which can establish the convergence of a series which is only conditionally convergent is theorem 6.13
6.23
Manipulations with series
Series are 'infinite sums'. It would therefore be optimistic to expect to be able to manipulate them just like 'finite sums'. Indeed, only absolutely convergent series may be freely manipulated. If one tries to obtain results by manipulating divergent or conditionally convergent series, only nonsense can be expected in general.
6.24
Example Consider the following argument.
0=0+0+0+
...
(l  1) + (1 1) + (l  1) + ... =
11+11+11+
...
1 + ( 1 + 1) + ( 1 + 1) + ( 1 + 1) + ... =
1+0+0+0+
=
1.
...
62
Series
The error is not hard to find. The series divergent.
6.25
1 1+ 1 1+ 1 1+ ...
is
(Its terms do not tend to zero.)
Example
We know that the series
1_1+1_1+1_1+1_ 2 3 4 5 6 '7
...
converges conditionally. Denote its sum by s. We rearrange the terms of the series appear and obtain
1~!+~tt+?fob+~
the order in which
...
Let the nth partial sum of this series be t; Then
t3n =
1 1 1 1 1+... +2
4
2n  1
3
k
(1+ + ... + 2n (~+~+ 4 8
...
~
1 4n  2
1 4n
i
1) (~ + + ... + 4n
~
2)
+~) 4n
= (1+ ~ + ... + _1_) _ ~ (1+ ~ + ... + _1_) 3
2n  1
2
2n  1
3
 ~ (~ + ~ + ... + ~) 2
= ~
2
2
2n
4
(1~+~~+ 2
""* ~s as n ""*
3
4
... +_1__ 2nl
~)
2n
00.
Notice that the finite sum for t3n can validly be rearranged in any way we like. Since t3n+1  t3n""* 0 as nr+ 00 and t3n+2  t3n""* 0 as n :+ 00, it follows that the rearranged series converges to which is not the same as the sum s of our original series (because s =1=0).
h
6.26
Exercise (1)
Using partial fractions,
L 00
n=1 n(n (2)
3n  2
+ 1)(n + 2)
prove that = 1.
If (an) and (bn> are two sequences
of positive terms and
Series
63
an
 .~ I as n + bn
00
where I =1=0, prove that the series
either both converge or both diverge. Discuss the convergence or divergence of the series 1
00
2
00
L n=1 2n 
(ii)
(iii)
1
L
n=1
2'
+3
n
(3) Suppose that (an) is a decreasing sequence of positive terms such that ~;;' = 1 an converges. Prove that na; + as n + 00. [Hint: consider an+l
°
+ an+2 + ... + a2n']
(4) Let (rn) denote the rational numbers from (0,1) arranged in the sequence whose first few terms are t ~,1,1,t!' ...Prove that the series
diverges. (5) Determine whether or not the following series converge or diverge.
L n1 00
(ii)
(iv)
(
,)2
~5n (2n)!
f: (~r2 4 n + 1)
00
n
(v)
n=1
L
n=1
1
h/(n+l)yn} n
(6) If the sum of the conditionally convergent series
1!+!!+!
...
is s, prove that the sum of the rearranged series
1+! ! +! + ~! + ~+ h i + ... is h· [ Hint: t3n =
11
1+ 3''2+'"
+ 4n
1 3
1
+ 4n 1 
1] Ln '
7
FUNCTIONS
7.1
Notation
A [unction [ from a set A to a set B (write f : A + B) defines a rule which assigns to each x E A a unique element y E B. The element y is called the image of the element x and we write y = [(x). When A and B are sets of real numbers we can draw the graph of the function as in the diagram below. The defining property of a function ensures that each vertical line drawn through a point of A cuts the graph in one and only one place. I
I
+' 0, we can find
a0
> 0 such
that
If(x) /1 < € provided that b  0
< x < b.
The number If(x) II is the distance betweenf(x) and I. We can think of it as the error in approximating to / by f(x). The definition of the statement f(x) ~ I as x ~ b  then amounts to the assertion that we can make the error in approximating to I by f(x) as small as we like by taking x sufficiently close to b on the left.
8.2
Limits from the right
Suppose again that f is defined on an interval (a, b). We say that f(x) tends (or converges) to a limit I as x tends to a from the right and write
f(x) ~ 1 as x ~ a +., or, alternatively, x
Urn r+
a+
f(x)
= I
/ I I
I
I
if the following criterion is satisfied. 74
a
75
Limits of [unctions
Given any I[(x) II
€
> 0, we can find
> 0 such that
0, we can find
a8
> 0 such that
I[(x)·/I' 0, we have found a 8 > 0 such that If(x)  f(~)1 < e provided that I x  ~ 1< 8. Hence f(x) ~ fen as x ~ ~ and so fis continuous at the point ~. Note that we showed that I f(x)  f(~) I < e 'provided that I x  ~ i < 8' instead of 'provided that 0 < lx  ~ 1< 8 as appears in the definition of f(x) ~ I as x ~~. This is because, when x =~,it is automatically true that I I(x)  f(~) I < e because then I f(x)  f(~) 1= o. It is only when we are considering a limit 1=1= fen that it is important to exclude the possibility that x = ~.
8.8
Connexion with convergent sequences
Theorem Letfbe defined on (a, b) except possit ly for ~ E (a, b). Then fex) ~ I as x ~ ~ if and only if, for each sequence (xn) of points of (a, b) such that Xn =1= ~ (n = 1,2, ... ) and Xn ~ ~ as n ~ it is true that f(xn) ~ I as n ~ 8.9
t'
00,
t, I,
00.
Proof (i) Suppose first that f(x) ~ I as x ~~. Let e > 0 be given. Then we can find a 8 > 0 such that If(x) /I < e provided that 0 < [x  ~I< 8. Now suppose that (xn) is a sequence of points of (a, b) such that x., if.: ~ (n = 1,2, ... ) and xn ~ ~ as n ~ Since 8 > 0 we can find an N such that for any n > N, IXn  ~I< 8. But x.; =1= ~ (n = 1,2, ... ) and so 0 < IXn  ~I< 8. But this implies that 00.
Given any e > 0, we have found an N such that, for any n > N, If(xn) II < e. Thus f(xn) ~ I as n ~ (ii) Now suppose that, for each sequence (xn) of points of (a, b) such that Xn =1= ~ = 1,2, ... ) and x.; ~ ~ as n ~ it is true that f(;en) ~ I as n ~ We now suppose that it is not true that f(x) ~ I as x ~ ~ and seek a contradiction. If it is not true that f(x) ~ I as x ~ ~ then, for some e > 0, it must be true that for each value of 8 > 0 we can find an x satisfying 0 < [x  ~I< 8 such that 00,
en
00,
If(x)
00.
II ;;,e.
In particular, if 8 such that
=
1/n, then we can find an Xn satisfying 0
< IXn 
~I< 1/n
If(xn) /1 ;;,e. But then (xn) is a sequence of points of (a, b) such that x.; =1= ~ (n = 1,2, ... ) and Xn ~ ~ as n ~ but for which it is not true that f(xn) ~ I as n ~ This contradicts our assumption above. 00
00.
Similar results to theorem 8.9 hold for convergence from the left or right. For example,f(x) ~ I as x ~ b  if and only if, for each sequence (xn) of points of (a, b) such that x.; ~ bas n ~ it is true that f(xn) ~ I as n ~ 00,
00.
Limits of functions
79
8.10
Example We show how theorem 8.9 can be used to prove that the function f: III 4> IR defined by f(x) = ox + ~ is continuous at every point ~ (see example 8.7). Let (xn> be any sequence of real numbers such that Xn 4> ~ as n 4> 00. By proposition 4.8, c\:xn + ~4> C\:~ + ~ as n 4> 00. From theorem 8.9 it follows that ox T ~ 4> C\:~ + ~ as x 4> ~ and this concludes the proof. y [(Xn)
= [(x) = ax + ~
       
[(0
8.11
Properties oflimits
The following propositions are analogues of the combination theorem and the sandwich theorem of chapter 4. They are easily deduced from these results with the help of theorem 8.9.
8.12
Proposition Letfandgbe defined on an interval (a, b) except possibly at ~ E (a, b). Suppose that f(x) 4> I as x 4> ~ and g(x) 4> m as x 4> ~ and suppose that A and J1are any real numbers. Then (i) A.f(x) + J,Jg(x) 4> Al + J1m as x 4> ~ (ii) f(x)g(x) 4> 1m as x 4> ~ (iii) f(x)/g(x) 4> 11m as x 4> ~ (provided m 0).
*
An important consequence of proposition 8.12 is the following result which we quote as a theorem.
8.13
Theorem A polynomial is continuous at every point. A rational function is continuous at every point at which it is defined.
Proof Example 8.7 shows that x proposition 8.12(ii) then shows that x" 4> Hence, if
as x 4>~. Repeated application of as x 4> ~ for each kEN.
4> ~
~h
P(x) = a.x" + an_1xn1 + ... + alx + aQ,
pen
then a repeated application of proposition 8.12(i) yields P(x) 4> as x 4> i.e. P is continuous at ~' To show that a rational function is continuous wherever it is defined we appeal to proposition 8 .1'2(iii).
t
80
Limits of functions
8.14 Proposition (sandwich theorem) Letf,g and h be defined on (a, b) except possibly at ~ E (a, b). Suppose that g(x) ~ I as x ~~, hex) ~ I as x ~ ~ and that g(x) ~ f(x) ~ h (x) except possibly when x = ~.Then f(x) ~ I as x ~ ~. Note The analogues of propositions 8.l2 and 8.l4 for convergence from the left or right are both true. ,t
I,
8.15
Exercise (1) Calculate the following limits (i) lim x+llx
Ix: + 44)
(ii) lim x+0l
Ix
+ ~X42 + 9) 3x +7
73
(2) Let f: IR~ IRbe defined by 3
f(x)
=
x
(x> 1)
1
(x
=
1 2x
(x
x I).
1)
Show thatf(x) ~ 2 as x ~ 1 andf(x) ~ 2 as x ~ 1 + using only the definitions of §8.l and §8.2. Deduce that the limit lim f(x)
x+l I
l. , I
I,
exists but is not equal to f(1). Draw a graph. (3) Let fbe defined for all x except x = 0 by f(x)
,
I'
.1;1
,,,
,
=
(1+x)2l x
(x
:f= 0).
Prove that/ex) ~ 2 asx ~ 0 even though/CO) is not defined. (4) Use the definition to show that Ix  ~I ~ 0 as x ~~. A functionfis defined on an open interval which contains ~ and, for each x in this interval,[(x) lies between ~ and x. Prove that/ex) ~ ~ as x ~~. (5) Let n EN. Prove that the function f: [0, 00) ~ IR defined by f(x)
=
x1/n
(x;;;': 0)
is continuous at each ~ > 0 and continuous on the right at O. [Hint: Use the sandwich theorem and exercise 3.11 (2) to prove that / is continuous at ~ > O. For continuity on the right at 0, appeal to the definition of [(x) ~ 0 as x ~ 0 +.] (6) Suppose thatf(x) ~ I as x ~~. If I> 0, show that, for some h >0,
Limits of [unctions f(x)
>0
provided that ~  h (ii) 1= O?
8.16
81
< x < ~+ h and x =I=~. What happens
if (i) 1 0 be given. Since fey) ~ I asy ~ 17, we can find a fl. > 0 such that If(y) II < € provided that 0 < Iy171 < fl.. Writingy =: g(x), we obtain If(g(x))
II
<e
(1)
provided that 0< Ig(x) 171
< fl.
But g(x) ~ 17 as x ~ ~ and /J. Ig(x) 171
(2)
> O. Hence we can find a 0 > 0 such that
< fl.
(3)
provided that (4) We would now like to say that (4) implies (3), which implies (2), which in turn, implies (1). This would complete the proof. But notice that (3) does not
82
I
I,
Limits of functions
imply (2) in general. We therefore need to use one of the hypotheses (i) or (ii) of the theorem. If we assume thatfis continuous at 1], then I = f(1]) and so IfCY)II < e even when y = 1] (see the remarks at the end of example 8.7). Thus, in this case, we may replace condition (2) by condition (3) and the argument described above goes through. If instead we assume hypothesis (ii), then we can be sure that g(x) 1] provided that x satisfies (4) for a sufficiently small value of 0 > 0. But then condition (3) can be replaced by condition (2) and the argument goes through again.
"*
8.18
Divergence
We write down two sample definitions. The reader will have little difficulty in supplying the definitions in other cases. We say that I(x) + + as x + ~ + if, given any H> 0, we can find a 8 > such that
°
00
1(X)>H provided that ~ <x < ~+ S, We say thatf(x) + I as x + that If(x) II < e provided that x

00
if, given any
find an X such
< X.
f
f(x)++ooasx>~
8.19
> 0, we can
€
~{
f(x)+{
+
Example Prove that
XI +
+
00
as x
+
0
asx>oo
+.
Proof Let H> 0 be given. We have to find a 0
> 0 such that
I >H
x
provided that 0
< x < o. Obviously
the choice 8 = Hl satisfies the requirement.
83
Limits of functions
8.20
Exercise (1) Prove the following (i) XI +  00 as x + 0(ii) x2 + + 00 as x + + 00. (2) Suppose that j'(y) + I asy + + 00 and g(x) + + 00 as x + + 00. Prove that f(g(x» + I as x + + 00. Explain why the difficulties encountered in theorem 8.17 do not occur in this case. (3) Suppose that fCY) + I as y + 0 +. Prove that f(x 1) + 1 as x + + 00. (4) Suppose thatfis defined on (a, b) except possibly for ~ E (a, b). Let <xn> be a sequence of points of (a, b) such that Xn (n = 1,2, ... ) and Xn + ~ as n + 00. Let be a sequence with the same properties. If 1'* m andf(xn) + I as n + 00 butf(yn) + mas n + 00, show that
'* ~
lirn [(x)
x+t
does not exist. [Hint: theorem 8.9] (5) Let f: IR + IR be defined by 1
f(x)
= (0
(x rational) (.x irrational).
Show that lirn f(x)
x+O
does not exist. (6) If fis defined as in the previous question, prove that lim {xf(x)}
x+O
= O.
9
CONTINUITY
9.1
Continuity on an interval
A function I is continuous at a point ~ if and only if I(x) l' I(~) as x ~ ~ (see §8.6). Geometrically, this means that the graph of I does not have a 'break' or 'jump' above the point ~. To say that I is continuous on an interval I should mean that its graph is unbroken above the interval I, i.e. that we can draw the part of the graph which lies above I without lifting our pencil from the paper. A precise mathematical definition is required. This depends on the type of interval in question.
A function I is continuous on an open interval I if and o~ is continuous at each point of I. ~
Where an endpoint of an interval belongs to the interval, a slightly more complicated definition is required. The most important case is that of a compact interval [a, b]. A function I is continuous on a compact interval [a, b] if and only if it is continuous at each point of (a, b) and continuous on the right at a and on the left at b.
,,'\; I
'f'
:
,
,,
I
84
W
I
"~/ b
f
I
continuous on (a,b)
)
I
, ,
,,
I~
I
I
I
a
b
f
continuous on [a,b J
85
Continuity
Note that, in both diagrams, the graph of I is unbroken above the interval in question. In general, a function I is continuous on an interval I provided that: (i) it is continuous at each point of I which is not an endpoint; (ii) it is continuous on the right at the left hand endpoint of I, if this exists and belongs to I; (iii) it is continuous on the left at the right hand endpoint of I, if this exists and belongs to I.
Examples
9.2
(i) All polynomials are continuous on every interval. All rational functions are continuous on any interval not containing a zero of the denominator (theorem 8.13). ' (ii) Let I: (0,00) + IR be defined by I(x) = l/x. Then I is continuous on the interval (0, 00) because H is continuous at each point ~ satisfying ~ > O. (Hi) Let I: IR + IR be defined by
I(x)
:=
t
X2
(x ~ 1)
3/2
(x> 1).
Then I is continuous at every point except x = 1. But I is continuous on the left at x = 1 and thus I is continuous on the interval [0, 1]. It is not continuous on [1,2] .
..1. 2
~ [0,1)
9.3 Proposition Let I be defined on an interval I. Then I is continuous on I if and only if, given any x E I and any € > 0, we can find a 8> 0 such that
II(x)  l(y)1
1)
x2+4
(v) f(x)
=2
x 4
(iv) f(x)
=
[1 0
(vi) f(x)
(O~.x~l) (otherwise)
X(Xl) =
[o
(O<x 0 and n EN, prove that 1
D{xlln}
=  xl/nxl. n
[Hint: exercise 3.11(6). Recall that the continuity of the nth root function was considered in exercise 8.15(5).] (2) Use the previous question and the rule for differentiating a composite function (theorem 10.13) to show that, for any rational number r, D{xr}
= rx'