1550251515-Classical_Complex_Analysis__Gonzalez_

202 Chapter4

for n > N., arbitrary p, and all z E E. Then the sufficient part of the

Cauchy condition yields the absolute and uniform convergence of L un(z)

on E.

Remark The series L kMn is said to be a majorant of the series L un(z)

Example Consider the Dirichlet series
00
1 1 1 1
L nz = p + 2z + · · · + nz + · · ·
n=l
where by nz we mean the principal value, i.e., nz = ezlnn so that lnzl =

e'" ln n = n'". Let 8 > 0 be an arbitrary but fixed positive number. Then

on the half-plane E = { z: x = Re z ;:::: 1 + 8} we have

I :z I = :x :; n1~8

But the numerical series L n-(lH) is known to converge. Hence, by the
Weierstrass M-test, the given series converges absolutely and uniformly
on E.
It is clear that L n-z converges absolutely at each point of E 1 =

{z: x = Rez > l} since for any given point z 1 E E 1 we may choose

0 < 8 < x1 - 1. However, the region of uniform convergence does not

extend to the whole half-plane Re z > 1.

Corollary 4.5 Let L~=o un(z), z E D, be a series of functions, and
suppose that

I

Un+ 1 ( z) I < r < l

un(z) -

( r a constant)

for all values of z E D1 C D, and for all values of n. In addition, suppose
that

luo(z)I:; K ( K a constant)

for all z E D1. Then the series converges absolutely and uniformly in D 1.

Proof By repeated application of the first inequality, we find that

lun(z)I:; rlun-1(z)I:; r^2 lun-2(z)I:; · · ·:; rnluo(z)I

so that

lun(z)I:; Krn

in view of the second inequality. The conclusion follows from the
Weierstrass M-test, since L Krn converges.