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as j ’ ∞. This contradiction proves the theorem.
Exercise 4.5.6. Use Theorem 4.5.5 to prove that a real-valued continuous
function on a closed bounded set is bounded. Use the trick given in Exer-
cise 4.3.5 (ii) to deduce that a real-valued continuous function on a closed
bounded set is bounded and attains its bounds (Theorem 4.3.4).
Exercise 4.5.7. If we work in Q rather than R use the function of Exam-
ple 1.1.3 to give an example of a function f : Q ’ Q which is continuous
but not uniformly continuous on [0, 2]. At which stage does our proof Theo-
rem 4.5.5 break down?
We shall use the fact that a continuous function on a closed bounded set
is uniformly continuous when we show that every continuous function on a
closed interval is integrable (Theorem 8.3.1) and it was in this context that
the notion of uniformity ¬rst arose.


4.6 The general principle of convergence
We know that an increasing sequence of real numbers tends to a limit if and
only if the sequence is bounded above. In this section we consider a similarly
useful result which applies to sequences in Rn .
67
Please send corrections however trivial to twk@dpmms.cam.ac.uk

De¬nition 4.6.1. We say that a sequence of points xn ∈ Rm is a Cauchy
sequence if, given any > 0, we can ¬nd n0 ( ) such that xp ’ xq < for
all p, q ≥ n0 ( ).
Our ¬rst lemma is merely8 algebraic.
Lemma 4.6.2. Any convergent sequence in Rm forms a Cauchy sequence.
Proof. Suppose that xn ’ x. Let > 0. By de¬nition, we can ¬nd an n0 ( )
such that

xn ’ x < /2 for all n ≥ n0 ( ).

We observe that, by the triangle inequality,

xp ’ xq ¤ xp ’ x + xq ’ x < /2 + /2 =

for all p, q ≥ n0 ( ), so we are done.
The converse to Lemma 4.6.2 is a powerful theorem of analysis.
Theorem 4.6.3. Any Cauchy sequence in Rm converges.
Proof. Suppose that xn ∈ Rm is a Cauchy sequence. Then, by de¬nition,
given any > 0, we can ¬nd N ( ) such that xp ’ xq < for all p, q ≥ N ( ).
In particular, if n ≥ N (1), we have

xn ¤ xn ’ xN (1) + xN (1) < 1 + xN (1) .

It follows that

xn ¤ max xr + 1
1¤r¤N (1)

for all n and so the sequence xn is bounded.
By the Bolzano-Weierstrass theorem, it follows that there is an x ∈ Rm
and a sequence n(1) < n(2) < . . . such that xn(j) ’ x as j ’ ∞. Thus,
given any > 0, we can ¬nd J( ) such that xn(j) ’ x < for all j ≥ J( ).
We now observe that if n ≥ N ( /2) and we choose a j such that j ≥
J( /2) and n(j) > N ( /2), then

xn ’ x ¤ xn ’ xn(j) + xn(j) ’ x < /2 + /2 = .

Thus xn ’ x as n ’ ∞ and we are done.
8
Remember that, from the standpoint of this book any argument, however di¬cult and
complicated it may be that does not involve the fundamental axiom is ˜mere algebra™.
68 A COMPANION TO ANALYSIS

Exercise 4.6.4. Show that, if any subsequence of a Cauchy sequence con-
verges, then the Cauchy sequence converges.
Combining Theorem 4.6.3 with Lemma 4.6.2, we get the general principle
of convergence.
Theorem 4.6.5. (General principle of convergence.) A sequence in
Rm converges if and only if it is a Cauchy sequence.
The general principle of convergence is used in the study of in¬nite sums.
De¬nition 4.6.6. If aj ∈ Rm we say that ∞ aj converges to s if
j=1

N
aj ’ s
j=1


as N ’ ∞. We write ∞ aj = s.
j=1
N
j=1 aj does not tend to a limit as N ’ ∞, we say that the sum
If

j=1 aj diverges.

[The de¬nition just given corresponds to how mathematicians talk but is
not how logicians talk. Formally speaking, ∞ aj either exists or it does
j=1
not and cannot converge or diverge. In an e¬ort to get round this some books
use locutions like ˜the series aj converges to the sum ∞ aj ™.]
j=1
We note that any result on sums can be rewritten as a result on sequences
and vice versa.
Exercise 4.6.7. We work in Rm .
(i) If sn = n aj , then ∞ aj converges to s if and only if sn ’ s as
j=1 j=1
n ’ ∞.
(ii) If we write s0 = 0 and aj = sj ’ sj’1 , then sn ’ s as n ’ ∞ if and
only if ∞ aj converges to s.
j=1

Exercise 4.6.8. By writing
N N N
(aj + bj ) = aj + bj ,
j=1 j=1 j=1


show that, if ∞ aj and ∞ ∞
bj converge, then so does j=1 (aj + bj ) and
j=1 j=1
that, in this case,
∞ ∞ ∞
(aj + bj ) = aj + bj .
j=1 j=1 j=1
69
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Exercise 4.6.9. If aj = bj for all j ≥ M and ∞ bj converges, show that
j=1

aj converges. Thus ˜the ¬rst few terms of an in¬nite sum do not a¬ect
j=1
its convergence™.

Exercise 4.6.10. Prove the general principle of convergence for sums, that
is to say, prove the following result. The sum ∞ aj converges if and only
j=1
q
if, given any > 0, we can ¬nd n0 ( ) such that j=p aj < for all
q ≥ p ≥ n0 ( ).

’ 0 as
Exercise 4.6.11. (i) Show that if j=1 aj converges, then an
n ’ ∞.
n ’1/2
≥ n1/2 , or otherwise, show that the
(ii) By observing that j=1 j
converse of part (i) is false.
∞ ’1+·
(iii) Show that, if · > 0, then j=1 j diverges. (We will obtain
stronger results in Exercises 5.1.9 and 5.1.10.)

We use the general principle of convergence to prove a simple but impor-
tant result.
Theorem 4.6.12. Let an ∈ Rm for each n. If ∞ aj converges, then
j=1
so does ∞ aj .
j=1

Proof. Since ∞ aj converges, the (trivial, algebraic) necessity part of
j=1
the general principle of convergence tells that, given any > 0, we can ¬nd
n0 ( ) such that q for all q ≥ p ≥ n0 ( ). The triangle inequality
j=p aj <
now tells us that
q q
aj ¤ aj <
j=p j=p


for all q ≥ p ≥ n0 ( ) and the (profound, analytic) su¬ciency part of the
general principle of convergence tells that ∞ aj converges.
j=1

Theorem 4.6.12 is often stated as saying that ˜absolute convergence implies
convergence™. We formalise this by making a de¬nition.

De¬nition 4.6.13. If aj ∈ Rm , we say that the sum aj is absolutely
j=1
convergent if ∞ aj converges.
j=1

Theorem 4.6.12 has the natural interpretation that if we wander around
R taking a sequence of steps of ¬nite total length then we must converge
m

on some point. However, although the result appears ˜intuitively evident™,
the proof ultimately depends on the fundamental axiom of analysis.
70 A COMPANION TO ANALYSIS

Exercise 4.6.14. The ¬rst proof of Theorem 4.6.12 that I was given went
as follows.
(i) We work ¬rst in R. Let an ∈ R for each n. If ∞ |aj | converges,
n=1
then set

a+ = a n , a ’ = 0 if an ≥ 0
n n

+
an = 0, an = ’an otherwise.

Use the fact that N a+ is an increasing sequence to show that limN ’∞ N a+
n=1 n n=1 n
N ’
exists. Show similarly that limN ’∞ n=1 an exists. Establish the equality
N N N
a’
a+ ’
an = n n
n=1 n=1 n=1


and use it to show that limN ’∞ N an exists.
n=1
(ii) By taking real and imaginary parts, use (i) to prove that if an ∈ C
for each n and ∞ |aj | converges, then so does ∞ aj .
j=1 j=1
(iii) Use (i) to prove Theorem 4.6.12.
Although this is, if anything, an easier proof than the one I gave, it is
a bit inelegant and does not generalise to to the context of general normed
spaces as discussed in Chapter 10.

Theorem 4.6.12 is often used in tandem with another simple but powerful
tool.

Theorem 4.6.15. (The comparison test.) We work in R. Suppose that
0 ¤ bj ¤ aj for all j. Then, if ∞ aj converges, so does ∞ bj .
j=1 j=1

Proof. Since An = n aj is an increasing sequence tending to a limit A we
j=1
know that An ¤ A for each n. Thus
n n
bj ¤ aj ¤ A
j=1 j=1


for each n and Bn = n bj is an increasing sequence bounded above. By
j=1
the fundamental axiom, Bn tends to a limit and the result follows.

Exercise 4.6.16. We work in R. Spell out in detail the proof that, if A1 ¤
A2 ¤ . . . and An ’ A as n ’ ∞, then Aj ¤ A for all j.

Exercise 4.6.17. (This should be routine for the reader. It will be needed
in the proof of Lemma 4.6.18.) We work in R.
71
Please send corrections however trivial to twk@dpmms.cam.ac.uk

(i) Show that n rj = (1 ’ r n+1 )/(1 ’ r).
j=0
∞ j
j=0 r converges if and only if |r| < 1 and that, if
(ii) Deduce that
|r| < 1, then

1
rj = .
1’r
j=0


We are now in a position to discuss power series ∞ an z n in C. The
n=0
reader may object that we have not discussed convergence in C but from
our point of view C is just R2 with additional algebraic structure. (Alterna-
tively, the reader should be able to supply all the appropriate de¬nitions and
generalizations for herself.)

Lemma 4.6.18. Suppose that an ∈ C. If n
n=0 an z0 converges for some
z0 ∈ C, then ∞ an z n converges for all z ∈ C with |z| < |z0 |.
n=0

Proof. Since ∞ an z0 converges, the general principle of convergence tells
n
n=0
n
us that |an z0 | ’ 0 as n ’ ∞. In particular, we can ¬nd an N such that
n n
|an z0 | ¤ 1 for all n ≥ N . Thus, setting M = 1 + max0¤n¤N |an z0 |, we have
n
|an z0 | ¤ M for all n ≥ 0.
Now suppose that |z| < |z0 |. We observe that

|z|n
n n
¤ M rn
|an z | = |an z0 | n
|z0 |

where r = |z|/|z0 |. Since 0 ¤ r < 1, the geometric sum ∞ M rn converges.
n=0
n n
Since 0 ¤ |an z | ¤ M r the comparison test (Theorem 4.6.15) tells us that
∞ n
n=0 |an z | converges. But, by Theorem 4.6.12, absolute convergence im-
plies convergence, so ∞ an z n converges.
n=0

Theorem 4.6.19. Suppose that an ∈ C. Then either ∞ an z n converges
n=0
for all z ∈ C, or there exists a real number R with R ≥ 0 such that
(i) ∞ an z n converges if |z| < R,
n=0
(ii) ∞ an z n diverges if |z| > R.
n=0
∞ n
We call R the radius of convergence of the power series n=0 an z . If
∞ n
n=0 an z converges for all z, then, by convention, we say that the power
series has in¬nite radius of convergence and write R = ∞. It should be no-
ticed that the theorem says nothing about what happens when |z| = R. In
Example 5.2.3 we shall see that the power series may converge for all such
z, diverge for all such z or converge for some points z with |z| = R. but not
for others.
72 A COMPANION TO ANALYSIS

Proof of Theorem 4.6.19. If ∞ an z n converges for all z, then there is noth-
n=0
ing to prove. If this does not occur, there must exist a z1 ∈ C such that

an z1 diverges. By Lemma 4.6.18, it follows that ∞ an z n diverges
n
n=0 n=0
whenever |z| > |z1 | and so the set

an z n converges}
S = {|z| :
n=0


is a bounded subset of R. Since ∞ an 0n converges, S is non-empty and
n=0
so has a supremum (see Theorem 3.1.7) which we shall call R.
By the de¬nition of the supremum, ∞ an z n diverges if |z| > R, so (ii)
n=0
holds. Now suppose |z| < R. By the de¬nition of the supremum we can ¬nd
a z0 with |z| < |z0 | < R such that ∞ an z0 converges. Lemma 4.6.18 now
n
n=0
tells us that ∞ an z n converges and we have shown that (i) holds.
n=0

Exercise 4.6.20. (i) If a ∈ C ¬nd, with proof, the radius of convergence of
∞ nn
n=0 a z .
(ii) Find, with proof, the radius of convergence of ∞ nn z n .
n=0
(iii) Conclude that every possible value of R with R ≥ 0 or R = ∞ can
occur as a radius of convergence.

Exercise 4.6.21. If ∞ an z n has radius of convergence R and |z1 | > R,
n=0
n
show that the sequence |an z1 | is unbounded.
(For further exercises on the radius of convergence look at Exercises K.59
and K.57.)

Theorem 4.6.19 and its proof (including the steps like Lemma 4.6.18 which
lead up to it) are both beautiful and important. We shall return to it later
(for example in Lemma 11.5.8) and extract still more information from it.
Time spent thinking about it will not be wasted. (If you would like an
exercise on the proof look at Exercise K.58.)
We conclude this section and the chapter with a rather less important
discussion which the reader may choose to skip or skim. The general principle
of convergence is obviously a very strong principle. It is natural to ask
whether it is as strong as the fundamental axiom of analysis. The answer is
that it is almost as strong but not quite. The general principle of convergence
together with the axiom of Archimedes are equivalent to the fundamental
axiom of analysis.

Exercise 4.6.22. Suppose that F is an ordered ¬eld that satis¬es the axiom
of Archimedes.
73
Please send corrections however trivial to twk@dpmms.cam.ac.uk

(i) Show that, if xn is an increasing sequence bounded above, then, given
any positive integer q, there exists an N (q) such that 0 ¤ xn ’ xm < 1/q for
all n ≥ m ≥ N (q).
(ii) Deduce that any increasing sequence bounded above is a Cauchy se-
quence.
(iii) Deduce that, if F satis¬es the general principle of convergence, it
satis¬es the fundamental axiom of analysis.

Later, in Appendix G, as a by-product of more important work done in
Section 14.4, we shall obtain an example of an ordered ¬eld that satis¬es the
general principle of convergence but not the axiom of Archimedes (and so
not the fundamental axiom of analysis).
Chapter 5

Sums and suchlike ™

This chapter contains material on sums which could be left out of a stream-
lined course in analysis. Much of the material can be obtained as corollaries
of more advanced work. However, I believe that working through it will help
deepen the reader™s understanding of the processes of analysis.


Comparison tests ™
5.1
How can we tell if a sum ∞ an converges?
n=1
We have already seen two very useful tools in Theorem 4.6.12 and Theo-
rem 4.6.15, which we restate here.

Theorem 5.1.1. (Absolute convergence implies convergence.) Let
an ∈ Rm for each n. If ∞ aj converges then so does ∞ aj .
j=1 j=1

Theorem 5.1.2. (The comparison test.) We work in R. Suppose that
0 ¤ bj ¤ aj for all j. Then, if ∞ aj converges, so does ∞ bj .
j=1 j=1

Comparison with geometric series gives the ratio test.

Exercise 5.1.3. We work in R. Suppose that an is a sequence of non-zero
terms with an+1 /an ’ l as n ’ ∞.
(i) If |l| < 1, show that we can ¬nd an N such that |an+1 /an | < (1 + l)/2
for all n ≥ N . Deduce that we can ¬nd a constant K such that |an | ¤
K((1 + l)/2)n for all n ≥ 1. Conclude that ∞ an converges.
n=1
(ii) If |l| > 1, show that |an | ’ ∞ as n ’ ∞ and so, in particular,

n=1 an diverges.

We can extend the result of Exercise 5.1.3 by using absolute convergence
(Theorem 5.1.1).

75
76 A COMPANION TO ANALYSIS

Lemma 5.1.4. (The ratio test.) Suppose that an is a sequence of non-zero
terms in Rm with an+1 / an ’ l as n ’ ∞.
(i) If |l| < 1, then ∞ an converges.
n=1
(ii) If |l| > 1, then an ’ ∞ as n ’ ∞ and so, in particular, ∞ an
n=1
diverges.
Exercise 5.1.5. Prove Lemma 5.1.4. What can you say if an+1 / an ’
∞?

Exercise 5.1.6. If a2n = a2n’1 = 4’n , show that n=1 an converges but
an+1 /an does not tend to a limit. Give an example of a sequence bn with
bn > 0 such that ∞ bn diverges but bn+1 /bn does not tend to a limit.
n=1

Notice that lemma 5.1.4 says nothing about what happens when l = 1.
In Exercise 5.1.9 (ii) the reader is invited to show that, if l = 1, we may have
convergence or divergence.
The ratio test is a rather crude tool and the comparison test becomes
more e¬ective if we can use other convergent and divergent series besides the
geometric series. Cauchy™s condensation test provides a family of such series.
(However, most people use the integral comparison test which we obtain later
in Lemma 9.2.4, so the reader need not memorise the result.)
Exercise 5.1.7. We work in R. Suppose that an is a decreasing sequence of
positive numbers.
(i) Show that
2N +1 ’1
2 N a2 N ≥ an ≥ 2N a2N +1 .
n=2N


2N a2N converges to A, then
(ii) Deduce that, if N =0

M
am ¤ A
m=1

for all M . Explain why this implies that ∞ am converges.
m=1
∞ ∞
2 N a2 N .
(iii) Show similarly that if m=1 am converges, so does N =0

Tidying up a bit (remember that the convergence of an in¬nite sum is
not a¬ected by the ¬rst few terms), we obtain the following lemma.
Lemma 5.1.8. (Cauchy™s condensation test.) We work in R. If an
is a decreasing sequence of positive numbers, then ∞ an and ∞ 2n a2n
n=1 n=1
converge or diverge together.
77
Please send corrections however trivial to twk@dpmms.cam.ac.uk

The next two exercises use logarithms and powers. These will be formally
de¬ned later in Sections 5.6 and 5.7 but we shall run no risk of circularity if
we use them in exercises.

Exercise 5.1.9. (i) Show that ∞ n’± converges if ± > 1 and diverges if
n=1
± ¤ 1.
(ii) Give an example of a sequence of strictly positive numbers an with
an+1 /an ’ 1 such that ∞ an converges. Give an example of a sequence of
n=1
strictly positive numbers an with an+1 /an ’ 1 and an ’ 0 as n ’ ∞ such
that ∞ an diverges.
n=1
∞ ’1
(iii) We know from (i) that n=1 n diverges. Use the inequality of
Exercise 5.1.7 (ii) to give a rough estimate of the size of N required to give
N ’1
n=1 n > 100.

Exercise 5.1.10. (i) Show that ∞ n’1 (log n)’± converges if ± > 1 and
n=2
diverges if ± ¤ 1. Give a rough estimate of the size of N required to give
N ’1 ’1
n=2 n (log n) > 100.
(ii) Show that ∞ n’1 (log n)’1 (log log n)’± converges if ± > 1 and di-
n=3
verges if ± ¤ 1. Give a rough estimate of the size of N required to give
N ’1 ’1 ’1
n=3 n (log n) (log log n) > 100.
(iii) Write down and answer the appropriate part (iii) for this question 1 .

However, as the following exercise makes clear, whatever series we use for
a comparison test, there must be some other series for which the test fails.

Exercise 5.1.11. Suppose an is a sequence of positive real numbers such
that ∞ an converges. Show that we can ¬nd N (1) < N (2) < . . . such that
n=1

M
an < 4’j for all M ≥ N (j).
n=N (j)


If we now set bn = an for 1 ¤ n ¤ N (1) and bn = 2j an for N (j) + 1 < n ¤
N (j + 1) [j ≥ 1], show that bn /an ’ ∞ as n ’ ∞ but ∞ bn converges.
n=1
Suppose un is a sequence of positive real numbers such that ∞ un di-
n=1
verges. Show that we can ¬nd a sequence of positive real numbers vn such
that vn /un ’ 0 as n ’ ∞ but ∞ vn diverges.
n=1

1
Functions with this sort of growth play an important role in certain parts of mathe-
matics as is indicated by the riddle ˜What sound does a drowning number theorist make?™
with the answer ˜Log log log . . . ™.
78 A COMPANION TO ANALYSIS

Conditional convergence ™
5.2
If the in¬nite sum ∞ aj is convergent but not absolutely convergent, we
j=1
call it conditionally convergent2 .
The reader should know the following simple test for convergence.
Lemma 5.2.1. (Alternating series test.) We work in R. If we have a
decreasing sequence of positive numbers an with an ’ 0 as n ’ ∞, then
∞ j+1
j=1 (’1) aj converges.
Further

N
j+1
(’1)j+1 aj ¤ |aN +1 |
aj ’
(’1)
j=1 j=1

for all N ≥ 1.
The last sentence is sometimes expressed by saying ˜the error caused by
replacing a convergent in¬nite alternating sum of decreasing terms by the
sum of its ¬rst few terms is no greater than the absolute value of the ¬rst
term neglected™. We give the proof of Lemma 5.2.1 as an exercise.
Exercise 5.2.2. We work in R. Suppose that we have a decreasing se-
quence of positive numbers an (that is, an ≥ an+1 ≥ 0 for all n). Set
sm = m (’1)j+1 aj .
j=1
(i) Show that the s2n form an increasing sequence and the s2n’1 form a
decreasing sequence.
(ii) By writing

s2n = a1 ’ (a2 ’ a3 ) · · · ’ (a2n’2 ’ a2n’1 ) ’ a2n

show that s2n ¤ a1 . Show also that s2n’1 ≥ 0. Deduce that s2n ’ ±,
s2n’1 ’ β as n ’ ∞ for some ± and β.
(iii) Show that ± = β if and only if an ’ 0 as n ’ ∞.
(iv) Suppose now that an ’ 0 as n ’ ∞ and so ± = β = l. Show that
sn ’ l as n ’ ∞.
(v) Under the assumptions of (iv), show that s2n ¤ l ¤ s2n+1 and deduce
that |l ’ s2n | ¤ a2n+1 . Show similarly that |l ’ s2n’1 | ¤ a2n for all n.
(vi) Check that we have indeed proved Lemma 5.2.1.
We use Lemma 5.2.1 to give part (iii) in the following set of examples of
what can happen on the circle of convergence.
2
The terminology we use for in¬nite sums is not that which would be chosen if the
subject were developed today, but attempts at reforming established terminology usually
do more harm than good.
79
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Example 5.2.3. (i) Show that ∞ nz n has radius of convergence 1 and
n=1
that ∞ nz n diverges for all z with |z| = 1.
n=1
(ii) Show that ∞ n’2 z n has radius of convergence 1 and that ∞ n’2 z n
n=1 n=1
converges for all z with |z| = 1.
(iii) Show that ∞ n’1 z n has radius of convergence 1 and that ∞ n’1 z n
n=1 n=1
converges if z = ’1 and diverges if z = 1.

Proof. We leave the details to the reader. In each case the fact that the series
diverges for |z| > 1 and converges for |z| < 1 is easily established using the
ratio test.
Lemma 5.2.1 is a special case of a test due to Abel.

Lemma 5.2.4. (Abel™s test for convergence.) Suppose that aj ∈ Rm
and there exists a K such that
n
aj ¤ K
j=1

for all n ≥ 1. If »j is a decreasing sequence of real positive numbers with
»j ’ 0 as j ’ ∞, then ∞ »j aj converges.
j=1

Exercise 5.2.5. Take m = 1 and aj = (’1)j+1 in Lemma 5.2.4.
(i) Deduce the alternating series test (without the estimate for the error).
(ii) By taking »j = 1, show that the condition »j ’ 0 cannot be omitted
in Lemma 5.2.4.
(iii) By taking »j = (1 + (’1)j )/2j, show that the condition »j decreasing
cannot be omitted in Lemma 5.2.4.

We set out the proof of Lemma 5.2.4 as an exercise.

Exercise 5.2.6. (i) Suppose that aj ∈ Rm and »j ∈ R. We write
n
Sn = aj .
j=1

Show that
q q q
»j (Sj ’ Sj’1 ) = »q+1 Sq + (»j ’ »j+1 )Sj ’ »p Sp’1
»j a j =
j=p j=p j=p

for all q ≥ p ≥ 1. This useful trick is known as partial summation by analogy
with integration by parts. Note that S0 = 0 by convention.
80 A COMPANION TO ANALYSIS

(ii) Now suppose that the hypotheses of Lemma 5.2.4 hold. Show that
q q
»j aj ¤ »q+1 K + (»j ’ »j+1 )K + »p K = 2»p K,
j=p j=p


and use the general principle of convergence to show that »j aj con-
j=1
verges.
Abel™s test is well suited to the study of power series as can be seen in
part (ii) of the next exercise.
Exercise 5.2.7. Suppose that ± is real.
(i) Show that ∞ n± z n has radius of convergence 1 for all values of ±.
n=1
(ii) Show that, if ± ≥ 0, ∞ n± z n diverges (that is, fails to converge)
n=1
for all z with |z| = 1.
(iii) Show that, if ± < ’1, ∞ n± z n converges for all z with |z| = 1.
n=1
(iv) Show that, if ’1 ¤ ± < 0, ∞ n± z n converges for all z with |z| = 1
n=1
and z = 1, but diverges if z = 1.
(v) Identify the points where ∞ n’1 z 2n converges.
n=1

So far as I know, it remains an open problem3 to ¬nd a characterisation
of those sets E ⊆ {z ∈ C : |z| = 1} such that there exists a power series
∞ ∞
n n
n=1 an z convergent when z ∈ E
n=1 an z of radius of convergence 1 with
and divergent when |z| = 1 and z ∈ E.
/

It is important to remember that, if we evaluate a sum like n=1 an
which is convergent but not absolutely convergent, then, as Exercise 5.2.8
below makes clear, we are, in e¬ect, evaluating ˜∞ ’ ∞™ and the answer we
get will depend on the way we evaluate it.
Exercise 5.2.8. Let aj ∈ R. Write

a+ = a j , a ’ = 0 if aj ≥ 0
j j
a+ = 0, a’ = ’aj if aj < 0.
j j

(i) If both ∞ a+ and ∞ a’ converge, show that ∞ |aj | converges.
j=1 j j=1 j j=1
(ii) If exactly one of j=1 aj and j=1 aj converges, show that ∞ aj
∞ ∞ ’
+
j=1
diverges.
(iii) Show by means of examples that, if both ∞ a+ and ∞ a’ di-
j=1 j j=1 j

verge, then j=1 aj may converge or may diverge.
(iv) Show that if ∞ aj is convergent but not absolutely convergent, then
j=1
∞ ∞
both j=1 aj and j=1 a’ diverge.
+
j
3
See [31].
81
Please send corrections however trivial to twk@dpmms.cam.ac.uk

(Exercise K.68 contains some advice on testing for convergence. Like
most such advice it ceases to be helpful at the ¬rst point that you really
need it.)
In his ground-breaking papers on number theory, Dirichlet manipulated
conditionally convergent sums in very imaginative ways. However, he was
always extremely careful to justify such manipulations. To show why he took
such care he gave a very elegant speci¬c counterexample which is reproduced
in Exercise K.72. The following more general and extremely striking result
is due to Riemann.
Theorem 5.2.9. Let aj ∈ R. If ∞ aj is convergent but not absolutely
j=1
convergent, then given any l ∈ R we can ¬nd a permutation σ of N+ = {n ∈
Z : n ≥ 1} (that is a bijection σ : N+ ’ N+ ) with ∞ aσ(j) convergent to
j=1
l.
We give a proof in the next exercise.

Exercise 5.2.10. Suppose that aj is convergent but not absolutely con-
j=1
vergent and that l ∈ R. Set

S(0) = {n ∈ N+ : an < 0}, T (0) = {n ∈ N+ : an ≥ 0}.

We use the following inductive construction.
n’1
At the nth step we proceed as follows. If j=1 aσ(j) < l, let mn =
min T (n) and set σ(n) = mn , T (n+1) = T (n)\{mn } and S(n+1) = S(n). If
n’1
j=1 aσ(j) ≥ l, let mn = min S(n) and set σ(n) = mn , S(n+1) = S(n)\{mn }
and T (n + 1) = T (n).
(i) Describe the construction in your own words. Carry out the ¬rst few
steps with an = (’1)n /n and suitable l (for example l = 2, l = ’1).
(ii) Show that σ is indeed a permutation.
(iii) If κ > |a| for all a ∈ S(n) ∪ T (n), m > n and S(m) = S(n),
T (m) = T (n) show that | m aσ(j) ’ l| < κ. (It may be helpful to consider
j=1
m
what happens when j=1 aσ(j) ’ l changes sign.)
(iv) Conclude that n aσ(j) ’ l as n ’ ∞.
j=1

Let me add that Example 5.2.9 is not intended to dissuade you from using
conditionally convergent sums, but merely to warn you to use them carefully.
As Dirichlet showed, it is possible to be both imaginative and rigorous.
In A Mathematician™s Miscellany [35], Littlewood presents an example
which helps understand what happens in Riemann™s theorem. Slightly dressed
up it runs as follows.
I have an in¬nite supply of gold sovereigns and you have none. At one
minute to noon I give you 10 sovereigns, at 1/2 minutes to noon I take one
82 A COMPANION TO ANALYSIS

sovereign back from you but give you 10 more sovereigns in exchange, at 1/3
minutes to noon I take one sovereign back from you but give you 10 more
sovereigns in exchange and so on (that is, at 1/n minutes to noon I take
one sovereign back from you but give you 10 more sovereigns in exchange
[n ≥ 2]). The process stops at noon. How rich will you be?
As it stands the question is not su¬ciently precise. Here are two refor-
mulations.
Reformulation A I have an in¬nite supply of gold sovereigns labelled s1 , s2 ,
. . . and you have none. At one minute to noon I give you the sovereigns
sj with 1 ¤ j ¤ 10, at 1/2 minutes to noon I take the sovereign s1 back
from you but give you the sovereigns sj with 11 ¤ j ¤ 20, in exchange,
at 1/3 minutes to noon I take the sovereign s2 back from you but give you
the sovereigns sj with 21 ¤ j ¤ 30, in exchange and so on (that is, at 1/n
minutes to noon I take the sovereign sn’1 back from you but give you the
sovereigns sj with 10n ’ 9 ¤ j ¤ 10n, in exchange). The process stops at
noon. How rich will you be? In this case, it is clear that I have taken all my
sovereigns back (I took sovereign sn back at 1/(n + 1) minutes to noon) so
you are no richer than before.
Reformulation B I have an in¬nite supply of gold sovereigns labelled s1 , s2 ,
. . . and you have none. At one minute to noon I give you the sovereigns
sj with 1 ¤ j ¤ 10, at 1/2 minutes to noon I take the sovereign s2 back
from you but give you the sovereigns sj with 11 ¤ j ¤ 20, in exchange, at
1/3 minutes to noon I take the sovereign s4 back from you but give you the
sovereigns the sj with 21 ¤ j ¤ 30, in exchange and so on (that is, at 1/n
minutes to noon I take the sovereign s2(n’1) back from you but give you the
sovereigns sj with 10n ’ 9 ¤ j ¤ 10n, in exchange). The process stops at
noon. How rich will you be? In this case, it is clear that I have given you
all my odd numbered sovereigns and taken none of them back. You are now
in¬nitely rich.
Exercise 5.2.11. Give a reformulation in which you end up with precisely
N sovereigns.
Remark: There is a faint smell of sulphur about Littlewood™s example. (What
happens if all the gold pieces are indistinguishable4 ?) However, most mathe-
maticians would agree that the original problem was not correctly posed and
that reformulations A and B are well de¬ned problems with the answers we
have given.
Littlewood™s paradox depends on my having an in¬nite supply of gold
sovereigns. In the same way Dirichlet showed that the phenomenon described
4
There is a connection with Zeno™s paradoxes. See Chapter II, Section 4 of [19] or in
rather less detail (but accompanied by many other splendid paradoxes) in [41].
83
Please send corrections however trivial to twk@dpmms.cam.ac.uk

in Example 5.2.9 cannot occur if the sum is absolutely convergent. We shall
prove this as Lemma 5.3.4 in the next section.


Interchanging limits ™
5.3
We often ¬nd ourselves wishing to interchange the order of two limiting
processes. Among the examples we shall discuss in the course of this book
are
∞ ∞ ∞ ∞
?
aij = aij (interchanging the order of summation)
i=1 j=1 j=1 i=1
b d d b
?
f (x, y) dx dy = f (x, y) dy dx (interchanging the order of integration)
a c c a
‚2f ? ‚2f
= (changing the order of partial di¬erentiation)
‚x‚y ‚y‚x
b b
?
lim fn (x) dx = lim fn (x) dx (limit of integral is integral of limit)
n’∞ a a n’∞
b b
d ‚f
?
f (x, y) dx = (x, y) dx (di¬erentiation under the integral sign)
dy ‚y
a a

(The list is not exhaustive. It is good practice to make a mental note each
time you meet a theorem dealing with the interchange of limits.)
Unfortunately, it is not always possible to interchange limits. The reader
should study the following example carefully. (A good counterexample may
be as informative as a good theorem.)

Exercise 5.3.1. Take anm = 1 if m ¤ n, anm = 0 otherwise. Write out the
values of anm in matrix form (say for 1 ¤ n, m ¤ 5). Find limn’∞ anm
when m is ¬xed and limm’∞ anm when n is ¬xed. Show that

lim ( lim anm ) = lim ( lim anm )
m’∞ n’∞ n’∞ m’∞

In his Course of Pure Mathematics [23], Hardy writes.

If L and L are two limit operations then the numbers LL z and
L Lz are not generally equal in the strict sense of the word ˜gen-
eral™. We can always, by the exercise of a little ingenuity, ¬nd z
so that LL z and L Lz shall di¬er from one another. But they are
equal generally if we use the word in a more practical sense, viz.
as meaning ˜in the great majority of such cases as are likely to
occur naturally™. In practice, a result obtained by assuming that
84 A COMPANION TO ANALYSIS

two limit operations are commutative is probably true; at any rate
it gives a valuable suggestion of the answer to the problem under
consideration. But an answer thus obtained, must in default of
further study . . . be regarded as suggested only and not proved.

To this I would add that, with experience, analysts learn that some limit
interchanges are much more likely to lead to trouble than others.
One particularly dangerous interchange is illustrated in the next example.

Exercise 5.3.2. Take anm = 1/n if m ¤ n, anm = 0 otherwise. Write out
the values of anm in matrix form (say for 1 ¤ n, m ¤ 5). Show that
∞ ∞
lim anm = lim anm .
n’∞ n’∞
m=1 m=1

We shall see a similar phenomenon in Example 11.4.12, and, in Exer-
cise 11.4.14 following that example, we shall look at a branch of mathematics
in which, contrary to Hardy™s dictum, the failure of limit interchange is the
rule rather than the exception.
Exercise 5.3.2 involves an ˜escape to in¬nity™ which is prevented by the
conditions of the next lemma.

Lemma 5.3.3. (Dominated convergence.) Suppose that cj ≥ 0 and

j=1 cj converges. If aj (n) ∈ R
m
and aj (n) ¤ cj for all n, then, if
aj (n) ’ aj as n ’ ∞, it follows that ∞ aj converges and
j=1

∞ ∞
aj (n) ’ aj
j=1 j=1

as n ’ ∞.

Proof. Let > 0 be given. Since j=1 cj converges, it follows from the
general principle of convergence that we can ¬nd an P ( ) such that
q
cj ¤ /3
j=p


for all q ≥ p ≥ P ( ). It follows that
q q
aj (n) ¤ cj ¤ /3 (1)
j=P ( )+1 j=P ( )+1
85
Please send corrections however trivial to twk@dpmms.cam.ac.uk

for all q ≥ P ( ) and all n. Allowing q to tend to in¬nity, we have

aj (n) ¤ /3 (2)
j=P ( )+1

for all n.
Returning to the inequality (1), but this time letting n tend to in¬nity
whilst keeping q ¬xed, we obtain
q q
aj ¤ cj ¤ /3
j=P ( )+1 j=P ( )+1


for all q ≥ P ( ). Since an increasing sequence bounded above converges,
∞ ∞
this shows that j=P ( )+1 aj . converges. Since j=P ( )+1 aj converges

absolutely it converges and the same must be true of j=P ( )+1 aj (n) We
now allow q to tend to in¬nity to obtain

aj ¤ /3. (3)
j=P ( )+1


Since aj (n) ’ aj , we can ¬nd Mj ( ) such that

aj (n) ’ aj ¤ /(6P ( ))

for all n ≥ Mj ( ). In particular, taking M ( ) = max1¤j¤P ( ) Mj ( ) we have

P( ) P( )
aj (n) ’ aj ¤ /3 (4)
j=1 j=1


for all n ≥ M ( ). Combining the inequalities (2), (3) and (4), we obtain
∞ ∞
aj (n) ’ aj ¤
j=1 j=1

for n ≥ M ( ) so we are done.
If we look at Littlewood™s sovereigns, the hypothesis in the dominated
convergence theorem can be interpreted as saying that we have only ∞ cjj=1
sovereigns to play with. It is thus, perhaps, not surprising that we can use
the dominated convergence theorem to prove that an absolutely convergent
sum can be rearranged in any way we wish without a¬ecting its convergence.
86 A COMPANION TO ANALYSIS

Lemma 5.3.4. We work in Rm . If ∞ aj is absolutely convergent and σ
j=1
is a permutation of N+ , then ∞ aσ(j) is absolutely convergent and
j=1
∞ ∞
aσ(j) = aj .
j=1 j=1

Proof. De¬ne aj (n) = aj if j ∈ {σ(1), σ(2), . . . , σ(n)} and aj (n) = 0
otherwise. Then aj (n) ¤ aj for all j and n and

n
aσ(j) = aj (n),
j=1 j=1

so, by the dominated convergence theorem,

n
aσ(j) ’ aj .
j=1 j=1

To see that ∞ aσ(j) is absolutely convergent, apply the result just obtained
j=1
with aj in place of aj .
The reader may feel that there is only one ˜natural™ method of de¬ning
r≥0 ar , to wit the standard de¬nition

N
ar = lim ar .
N ’∞
r=0
r≥0

I would disagree with her even for this case5 , but it is clear that there are
several ˜natural™ methods of de¬ning r,s≥0 ars . Among the possibilities are
N N N r
lim ars , lim a(r’s) s , and lim ars
N ’∞ N ’∞ N ’∞
r=0 s=0 r=0 s=0 r 2 +s2 ¤N

(that is, considering sums over squares, triangles or quadrants of circles).
Fortunately it is clear that each of these summation methods is a rearrange-
ment of the others and so, provided we have absolute convergence, it does not
matter which we use.
We give the details in the next lemma but the reader who is already
convinced can safely ignore them.
5
Communications engineers sometimes use ˜hard summation™

lim ar .
δ’0+
ar >δ
87
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Lemma 5.3.5. Suppose that E is an in¬nite set and that E1 , E2 , . . . and
F1 , F2 , . . . are ¬nite subsets of E such that
(i) E1 ⊆ E2 ⊆ . . . , ∞ Ek = E,
k=1
(ii) F1 ⊆ F2 ⊆ . . . , ∞ Fk = E.
k=1
Suppose further that ae ∈ Rm for each e ∈ E.
Then, if e∈EN ae tends to a limit as N ’ ∞, it follows that e∈EN ae ,
e∈FN ae tend to a limit as N ’ ∞ and
e∈FN ae and

lim ae = lim ae .
N ’∞ N ’∞
e∈EN e∈FN

Proof. Using condition (i) and the fact that the Xk are ¬nite, we can enu-
merate the elements of X as e1 , e2 , . . . in such a way that
EN = {e1 , e2 , . . . , eM (N ) }
for some M (N ). Thus
M (N )
a ej = ae
j=1 e∈EN

tends to a limit, so M (N ) aej is bounded by some K for all N and, since
j=1

all terms are positive, n j=1 aej is bounded by K for all n. Thus j=1 aej
converges absolutely.
We now observe that there is a permutation σ such that
P (N )
aeσ(j) = ae
j=1 e∈FN

for some integer P (N ). The required results now follow from Theorem 5.3.4.

Exercise 5.3.6. Explain how the previous lemma gives the following result
for ars ∈ Rm .
If any of the three limits
N N N r
lim ars , lim a(r’s)s , or lim ars
N ’∞ N ’∞ N ’∞
r=0 s=0 r=0 s=0 r 2 +s2 ¤N

exist, then so do the other two. Further, under this condition, the three limits
N N N r
lim ars , lim a(r’s)s , and lim ars
N ’∞ N ’∞ N ’∞
r=1 s=0 r=0 s=0 r 2 +s2 ¤N

exist and are equal.
88 A COMPANION TO ANALYSIS

There are two other ˜natural™ methods of de¬ning ars in common
r,s≥0
use, to wit
∞ ∞ ∞ ∞
ars and ars
r=0 s=0 s=0 r=0

or more explicitly
N M M N
lim lim ars and lim lim ars .
N ’∞ M ’∞ M ’∞ N ’∞
r=0 s=0 s=0 r=0

We shall show in the next two lemmas that, provided we have absolute con-
vergence, these summation methods are also equivalent.

Lemma 5.3.7. Let ars ∈ Rm for r, s ≥ 0. The following two statements
are equivalent.
(i) N N
s=0 ars tends to a limit as N ’ ∞.
r=0
(ii) ∞ ars converges for all r ≥ 0 and ∞ ∞
s=0 ars converges.
s=0 r=0

Further, if statements (i) and (ii) hold, then s=0 ars converges for all
r ≥ 0 and ∞ ∞
s=0 ars converges.
r=0

Proof. We ¬rst show that (i) implies (ii). Observe that, if M ≥ r, then
M M M N N
ars ¤ aus ¤ lim aus .
N ’∞
s=0 u=0 s=0 u=0 s=0


Thus, since an increasing sequence bounded above tends to a limit, ars
s=0
converges for all r ≥ 0. Now observe that if N ≥ M
M N N N
ars ¤ ars .
r=0 s=0 r=0 s=0


Thus, allowing N ’ ∞ whilst keeping M ¬xed, (remember from Lemma 4.1.9 (iv)
that the limit of a ¬nite sum is the sum of the limits) we have

M N N
ars ¤ lim ars .
N ’∞
r=0 s=0 r=0 s=0

∞ ∞
Thus, since an increasing sequence bounded above tends to a limit, ars
r=0 s=0
converges.
89
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Now we show that (ii) implies (i). This is easier since we need only note
that
∞ ∞ ∞
N N N
ars ¤ ars ¤ ars
r=0 s=0 r=0 s=0 r=0 s=0

and use, again, the fact that an increasing sequence bounded above tends to
a limit.
Finally we note that, if (ii) is true, the fact that absolute convergence
implies convergence immediately shows that ∞ ars converges for all r ≥ 0.
s=0
We also know that
∞ ∞
ars ¤ ars ,
s=0 s=0
∞ ∞
so the comparison test tells us that ars converges.
r=0 s=0

Lemma 5.3.8. Let ars ∈ Rm for r, s ≥ 0. If the equivalent statements (i)
and (ii) in Lemma 5.3.7 hold, then
∞ ∞ N N
ars = lim ars .
N ’∞
r=0 s=0 r=0 s=0

Proof. By Lemma 5.3.7 and 5.3.5, we know that ∞ ∞ N N
s=0 ars and limN ’∞ ars
r=0 r=0 s=0
exist. We need only prove them equal.
We prove this by using the dominated convergence theorem again. Set
ars (N ) = ars if 0 ¤ r ¤ N and 0 ¤ s ¤ N , and ars (N ) = 0 otherwise. If r
is ¬xed,
∞ ∞
N
ars ’
ars (N ) = ars .
s=0 s=0 s=0

But we know that
∞ ∞
N N
ars (N ) ¤ ars (N ) ¤ ars ¤ ars
s=0 s=0 s=0 s=0

and ∞ ∞
ars converges. Thus the dominated convergence theorem
r=0 s=0

br (N ) with br (N ) = ∞ ars (N ) gives
applied to r=0 s=0

∞ ∞ ∞ ∞
N N
ars (N ) ’
ars = ars
r=0 s=0 r=0 s=0 r=0 s=0

as required
90 A COMPANION TO ANALYSIS

We have proved the following useful theorem.

Lemma 5.3.9. (Fubini™s theorem for sums.) Let ars ∈ Rm . If any one
of the following three objects,
∞ ∞ ∞ ∞
N N
lim ars , ars , ars
N ’∞
r=0 s=0 r=0 s=0 s=0 r=0

is well de¬ned, they all are, as are
∞ ∞ ∞ ∞
N N
lim ars , ars , ars .
N ’∞
r=0 s=0 r=0 s=0 s=0 r=0

Further
∞ ∞ ∞ ∞
N N
lim ars , = ars , = ars .
N ’∞
r=0 s=0 r=0 s=0 s=0 r=0

This theorem allows us to interchange the order of summation of in¬nite
sums (provided we have absolute convergence). I attach the name Fubini to
it because Fubini proved a general and far reaching theorem of which this is
a special case.

Exercise 5.3.10. (i) Set arr = 1 for all r ≥ 1 and ar r’1 = ’1 for all r ≥ 2.
Set ars = 0 otherwise. Write out the matrix with entries ars where 1 ¤ r ¤ 4,
1 ¤ s ¤ 4. Show, by direct calculation, that
∞ ∞ ∞ ∞
ars = ars .
r=1 s=1 s=1 r=1

(ii) Set b11 = 1, b1s = 0 for s ≥ 2 and

brs = 2’r+2 for 2r’2 ¤ s ¤ 2r’1 ’ 1
brs = ’2’r+1 for 2r’1 ¤ s ¤ 2r ’ 1
brs = 0 otherwise,

when r ≥ 2. Write out the matrix with entries brs where 1 ¤ r ¤ 4, 1 ¤ s ¤
8. Show by direct calculation that
∞ ∞ ∞ ∞
brs = brs .
r=1 s=1 s=1 r=1
91
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Summing over triangles rather than squares and using Lemma 5.3.5, we
obtain another version of Fubini™s theorem which we shall use in the next
section.
∞ ∞
Lemma 5.3.11. If aij converges, then writing
i=0 j=0


sn = aij ,
i+j=n


we have n=0 sn absolutely convergent and
∞ ∞ ∞
sn = aij .
n=0 i=0 j=0


We are no longer so concerned with tracing everything back to the fun-
damental axiom but that does not mean that it ceases to play a fundamental
role.

Exercise 5.3.12. We work in Q. Show that we can ¬nd x0 , x1 , x2 , x3 , . . .
an increasing sequence such that x0 = 0 and xj ’ xj’1 ¤ 2’j for j ≥ 1, but
xn does not tend to a limit.
Set x0 = 0, a1j = xj ’ xj’1 , a2j = 2’j ’ a1j , and aij = 0 for i ≥ 3.
Show that aij ≥ 0 for all i, j, that ∞ aij exists for all j and ∞ ∞
i=1 aij
i=1 j=1

exists. Show, however, that j=1 aij does not exist when i = 1 or i = 2.


The exponential function ™
5.4
We have proved many deep and interesting theorems on the properties of
continuous and di¬erentiable functions. It is somewhat embarrassing to ob-
serve that, up to now, the only di¬erentiable (indeed the only continuous
functions) which we could lay our hands on were the polynomials and their
quotients. (Even these were obtained as an afterthought in Exercise 4.2.24.
We shall use results from that exercise in this section, but all such results
will be proved in a wider context in Chapter 6.) In the next four sections we
widen our repertoire considerably.
We start with the function exp. Historically, the exponential function
was developed in connection with Napier™s marvelous invention of the loga-
rithm as a calculating tool (see Exercise 5.6.6). However, if by some historic
freak, mathematics had reached the state it was in 1850 whilst bypassing the
exponential and logarithmic functions, the exponential function might have
been discovered as follows.
92 A COMPANION TO ANALYSIS

One way to obtain new functions is to look for solutions to di¬erential
equations. Consider one of the simplest such equations y (x) = y(x). With-
out worrying about rigour (˜In a storm I would burn six candles to St George
and half a dozen to his dragon™) we might try to ¬nd a solution in the form
of a power series y(x) = ∞ aj xj .
j=0

Plausible statement 5.4.1. The general solution of the equation

y (x) = y(x), ()

where y : R ’ R is a well behaved function, is

xj
y(x) = a
j!
j=0


with a ∈ R.

Plausible argument. We seek for a solution of of the form y(x) = ∞ aj xj .
j=0
Assuming that we can treat a power series in the same way as we can treat
a polynomial, we di¬erentiate term by term to obtain y (x) = ∞ jaj xj’1 .
j=1
Equation thus becomes
∞ ∞
aj x j = (j + 1)aj+1 xj
j=0 j=0


which may be rewritten as

(aj ’ (j + 1)aj+1 )xj = 0.
j=0


Now, if a polynomial P (x) vanishes for all values of x, all its coe¬cients are
zero. Assuming that the result remains true for power series, we obtain

aj ’ (j + 1)aj+1 = 0

for all j ≥ 0 and a simple induction gives aj = a0 /j!. Setting a = a0 , we
have the result.

Later in the book we shall justify all of the arguments used above (see Theo-
rem 11.5.11 and Exercise 11.5.13) but, for the moment, we just use them as
a heuristic tool.
We can make an immediate observation.
93
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Lemma 5.4.2. The power series

zj
j!
j=0

has in¬nite radius of convergence.
Proof. If z = 0, then, setting uj = z j /j!, we have |uj+1 |/|uj | = |z|/(j+1) ’ 0
as j ’ ∞. Thus, by the ratio test, ∞ z j /j! is absolutely convergent, and
j=0
so convergent, for all z.

j=0 z /j! for all z ∈ C. A little playing
j
We may thus de¬ne e(z) =
around with formulae would lead to the key observation.
Lemma 5.4.3. If z, w ∈ C, then
e(z)e(w) = e(z + w)
Proof. Observe that ∞ |z r |/r! converges to e(|z|) and ∞
|ws |/s! con-
r=0 s=0
verges to e(|w|). Thus, writing ars = (|z r |/r!)(|w s |/s!),
∞ ∞ ∞ ∞
|z r | |w|s
|ars | =
r! s!
r=0 s=0 r=0 s=0
∞ ∞
|z r | |w|s
=
r! s!
r=0 s=0

|z r |
= e(|w|) = e(|z|)e(|w|).
r!
r=0

Thus ∞ ∞
s=1 ars converges absolutely and, by Lemma 5.3.11, it follows
r=1
that, if we write
cn = ars ,
r+s=n

we have n=0 cn absolutely convergent and
∞ ∞ ∞
cn = aij .
n=1 i=1 j=1

But
n n
z r wn’r
cn = ar(n’r) =
r! (n ’ r)!
r=0 r=0
n
n r n’r (z + w)n
1
= zw = ,
n! r n!
r=0
94 A COMPANION TO ANALYSIS

and so
∞ ∞ ∞ ∞ ∞
zr ws
e(z)e(w) = = ars = cn = e(z + w).
r! s!
r=0 s=0 r=1 s=1 n=1




It should be noticed that the essential idea of the proof is contained in its
last sentence. The rest of the proof is devoted to showing that what ought
to work does, in fact, work.

Exercise 5.4.4. (Multiplication of power series.) The idea of the pre-
vious lemma can be generalised.
(i) Let »n , µn ∈ C and γn = n »j µn’j . Show, by using Lemma 5.3.11,
j=0
or otherwise, that, if ∞ »n and ∞ µn are absolutely convergent so is
n=0 n=0

γn and
n=0

∞ ∞ ∞
»n µn = γn .
n=0 n=0 n=0


(ii) Suppose that ∞ an z n has radius of convergence R and that ∞ bn z n
n=0 n=0

has radius of convergence S. Explain why, if |w| < min(R, S), both n=0 an wn
and ∞ bn wn converge absolutely. Deduce that, if we write cn = r+s=n ar bs
n=0
then ∞ cn wn converges absolutely and
n=0

∞ ∞ ∞
n n
cn w n .
an w bn w =
n=0 n=0 n=0

∞ n
Thus n=0 cn z has radius of convergence at least min(R, S) and, if
|w| < min(R, S), the formula just displayed applies. This result is usually
stated as ˜two power series can be multiplied within their smaller radius of
convergence™.

Exercise 5.4.5. Use Exercise 5.4.4 directly to prove Lemma 5.4.3.

z n has radius of convergence 1
Exercise 5.4.6. (i) Prove directly that n=0
and

1
zn =
1’z
n=0

for |z| < 1.
95
Please send corrections however trivial to twk@dpmms.cam.ac.uk

∞ n
(ii) Use Exercise 5.4.4 to show that n=0 (n + 1)z has radius of conver-
gence at least 1 and that

1
(n + 1)z n =
(1 ’ z)2
n=0

for |z| < 1. Show that ∞ (n + 1)z n has radius of convergence exactly 1.
n=0
(iii) It is trivial that 1 + z has radius of convergence ∞. Use this obser-
vation and others made earlier in the exercise to show that, for appropriate
choices of ∞ an z n and ∞ bn z n , we can have, in the notation of Exer-
n=0 n=0
cise 5.4.4,
(a) R = 1, S = ∞, ∞ cn z n has radius of convergence ∞.
n=0

(b) R = 1, S = ∞, n=0 cn z n has radius of convergence 1.
[See also Exercise K.234.]
The next exercise gives an algebraic interpretation of Lemma 5.4.3.
Exercise 5.4.7. Check that C is an Abelian group under addition. Show that
C \ {0} is an Abelian group under multiplication. Show that e : (C, +) ’
(C \ {0}, —) is a homomorphism.
The following estimate is frequently useful.
Lemma 5.4.8. If |z| < n/2, then
n’1
zj 2|z|n
e(z) ’ ¤ .
j! n!
j=0

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