Mathematical Methods for Physics and Engineering : A Comprehensive Guide

24.3 POWER SERIES IN A COMPLEX VARIABLE

This series is absolutely convergent if

∑∞

n=0

|an|rn, (24.12)

which is a series of positive real terms, is convergent. Thus tests for the absolute

convergence of real series can be used in the present context, and of these the

most appropriate form is based on the Cauchy root test. With theradius of

convergenceRdefined by

1

R

= lim

n→∞

|an|^1 /n, (24.13)

the series (24.10) is absolutely convergent if|z|<Rand divergent if|z|>R.

If|z|=Rthen no particular conclusion may be drawn, and this case must be

considered separately, as discussed in subsection 4.5.1.

A circle of radiusRcentred on the origin is called thecircle of convergence

of the series

∑

anzn.ThecasesR= 0 andR=∞correspond, respectively, to

convergence at the origin only and convergence everywhere. ForRfinite the

convergence occurs in a restricted part of thez-plane (the Argand diagram). For

a power series about a general pointz 0 , the circle of convergence is, of course,

centred on that point.

Find the parts of thez-plane for which the following series are convergent:

(i)

∑∞

n=0

zn

n!

, (ii)

∑∞

n=0

n!zn, (iii)

∑∞

n=1

zn

n

.

(i) Since (n!)^1 /nbehaves likenasn→∞we find lim(1/n!)^1 /n=0.HenceR=∞and the
series is convergent for allz. (ii) Correspondingly, lim(n!)^1 /n=∞. ThusR=0andthe
series converges only atz= 0. (iii) Asn→∞,(n)^1 /nhas a lower limit of 1 and hence
lim(1/n)^1 /n=1/1 = 1. Thus the series is absolutely convergent if the condition|z|<1is
satisfied.

Case (iii) in the above example provides a good illustration of the fact that

on its circle of convergence a power series may or may not converge. For this

particular series, the circle of convergence is|z|= 1, so let us consider the

convergence of the series at two different points on this circle. Takingz=1,the

series becomes
∑∞

n=1

1

n

=1+

1

2

+

1

3

+

1

4

+···,

which is easily shown to diverge (by, for example, grouping terms, as discussed in

subsection 4.3.2). Takingz=−1, however, the series is given by

∑∞

n=1

(−1)n

n

=−1+

1

2

−

1

3

+

1

4

−···,