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