1550251515-Classical_Complex_Analysis__Gonzalez_

Sequences, Series, and Special Functions 563

as N ---+ oo. Now, given e > 0, choose N (i.e., lzl sufficiently close to 1)

such that lnanl < e for n > N. Then, with 0 < lzl < 1, we have

f: anzn = I f: nan z: $ N: l f: lzln

n=N+l n=N+l n=N+l

f

< (N + 1)(1 - lzl) < e

since N + 1 > 1/(1 - lzl), or (N + 1)(1 - lzl) > 1.

On the other hand, we have

ll - znl = 1(1 -z)(l + z + .. · + zn-l )I S ll - zln

and if z E <J' the inequality 11-zl S k(ll -lzl) is satisfied for some constant

k > 2. Hence

t. an(l - zn)I $ t. lnan(l - z)I S k(l - lzl) t. nlanl

k N

SN I:nlanl

n=O

By Theorem 4.1-15, nlanl ---+ 0 implies that (1/N) L:~=O nlanl ---+ 0 as
N ---+ oo.
Consequently, we have

.l'.;. :•z" -t. a.( 1 - z•) I ,; .J;:+ :·z• I + t. a. (1 - z•) I < 2<

for N large enough, which proves the· theorem.

Remark J. E. Littlewood has shown [23] that the condition an= o(l/n)

in Tauber's theorem can be replaced by an = 0(1/n). For an alternative
proof of Littlewood's theorem, see [37].

Exercises 8.2

1. Show that the series

converges absolutely at all points of its circle of convergence. Investigate

whether or not z = 1 is a singular point of f.