1550251515-Classical_Complex_Analysis__Gonzalez_

Sequences and Series 205

The number R is called the radius of convergence of the power series, and

the circle jzj = R is called the circle of convergence. The last denomination

is a misnomer, since a power series may converge or diverge on jzj = R (if

0 < R < oo ). What is really meant is that the series converges absolutely

in the circular region (or disk) jzj < R, unless R = 0, in which case the

set of convergence reduces to a point, or R = oo, in which case the region

of convergence is the complex plane.

Proof (1) If R = 0 or limsup \/fan! = oo, then for any ]{ > 0 we have

\/Ian I > ]{ or Ian I > Kn for infinitely many values of n. Hence, for any

given zo -=/:= 0 we have

lanzo I > Kn lzo In > 1

for infinitely many n by choosing K > 1/lz 0 j. Therefore, anzo does not

tend to zero as n -t oo, and the series diverges. However, for z = 0 we

have Sn(O) = a 0 and the series converges (absolutely to ja 0 j).

(2) If R = oo or lim sup \/Ian I = 0, then for any € > 0 there is Ne such

that \/fan! < €or Ian! < En provided that n > N •. Hence, for any given

Zo-=/:= 0 we have lanzol < Enlzoln for n > N •. Choosing€ such that Elzol < 1,

we see that L:; anzo converges by comparison with the convergent geometric

series L:; Enlzoln. Since every power series converges absolutely for z = 0,

it follows that in this case the series converges absolutely for every z E C.

(3) Applying the third form of the root test [Section 4.9, part (b )] to

the series L:; anzn, and using ( 4.11-4), we have

, so that if lzl < R, the series L:; anzn converges absolutely, and it diverges
if lzl > R.

To see that a power series may converge or diverge at a point on jzl = R,

consider the series L:;;'°(l/n)zn for which R = 1. In this case the series

diverges for z = 1, while it converges for z = -1. In some cases the series

may converge at every point of jzl = R, while in others it may diverge

everywhere on the same circle (see Exercises 4.2, problems 21, 22, and 23).

To show that the series converges uniformly on jzl ::::; r if 0 < r < R,

choose a point z 1 such that r < lz 1 I < R. Then we have, for lzl ::::; r.

lanzn I ::::; Ian lrn < Ian llz1 In

Since the numerical series L:; lanllz1 In converges, by the Weierstrass M-test

the series L:; anzn converges uniformly (and absolutely) on lzl ::::; r.

( 4) Let z 0 be an arbitrary point in lz I < R. Choose r such that lzo I <

r < R. Since the power series converges uniformly on lzl ::::; r and each