1550251515-Classical_Complex_Analysis__Gonzalez_

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560

Hence

11-zl < k

1-lzl -

Chapters

for any z in (}' except the vertex A. Consequently, if we let z ~ 1 along

any path, ultimately in <J' we have IRn(z)I < e(l + k) for n large enough,

which means that

n

Sn(z) = l::>kzk

k=O

converges uniformly as z ~ 1, z E <J'. Then by Theorem 4.18 we have

(z = 1 being an accumulation point of <J')

00 00

l.im f(z) = li"m lim Sn(z) = lim '"""akzk = '"""ak = S

z---+1 z---+l n-+oo z-+I ~ L...J

zEu zEu zEu k=O k=O

Remarks Since f(l) = S, Abel's theorem means that f(z) is continuous

in a restricted sense at z = 1.

The more general case of a power series with radius of convergence R

(0 < R < oo) converging at a point z 0 =Reio on the circle of convergence

can be reduced to the case discussed in Theorem 8.15 by replacing z by z / zo.

The condition ll-zl/(1-lzl) :::; k needs to be imposed on ll-zl/(1-lzl)

since this expression becomes unbounded for z close to 1 yet much closer

to the circle lzl = 1. For instance, letting 0 < 11-zl = p < 1, 1 - lzl = p^2 ,

we have

----? 00

As a corollary of Theorem 8.15 we shall prove Abel's theorem con-

cerning the product of two converging series. See the Remark following

Theorem 4.14.

Corollary 8. 7 If two series I.: an and I.: bn are convergent with sums A

and B, respectively, and if the product series I.: en, formed according to

Cauchy's rule, is also convergent with sum C, then C =AB.

Proof Consider the power series

00 00 00

f(z) = L anzn, g(z) = L bnzn,

n=O n=O