1550251515-Classical_Complex_Analysis__Gonzalez_

558 Chapters

Proof As a preliminary step we shall prove the so-called Abel's formula of

summation by parts; namely, if {bn} and {en} are any two sequences, then

n+p n+p

L bk(ek - ek-1) = bn+pCn+p - bnen - L (bk - bk-1)ek-1 (8.7-1)

k=n+l k=n+l

In fact, we have

n+p n+p n+p

L bk(ek - ck-1) = L bkck - L bkek-1

f,=n+l k=n+l k=n+l

n+p-1 n+p

= bn+pen+p - bnen + L bkek - L bkck-1

k=n k=n-1-1

n+p n+p

= bn+pen+p - bnen + L bk-lek-1 - L bkek-1

k=n+l k=n+l

n+p

= bn+pen+p - bnen - L (bk - bk-1)ek-1

k=n+l

Now if we apply (8.7-1) to the partial remainder of the power series, that is,

n+p ·'

Rn,p(z) = .W ''""' akz k , n 2:: 0, p 2::^1

k=n+l

with bn == zn and Cn = l:~n+l ak, we obtain, noting that en -Cn-1 = -an,

n+p

Rn,p(z) = - L bk(ek - ek-1)

k=n+l

n+p

= - Cn+pzn+p + enZn + L (zk - Zk-l)ek-1 (8.7-2)

k=n+l

Since L: an converges, we have en ---+ 0 as n ---+ oo. Hence if we take Jzl < 1

and let .P ---+ oo, (8.7-2) gives
00 00
Rn(z) = L akzk = CnZn - (1-z) L zk-lek-l (8.7-3)
k=n+l k=n+l

For any given e > 0 there exists N such that n 2:: N implies Jeni < e.

Thus for n 2:: N we have

1-Rn(z)I < e(l + 11-zl ~ izlk-l) = e (1 + ~

1

_=-

1

;:) (8.7-4)