Mathematical Methods for Physics and Engineering - Matematica.NET

24.3 POWER SERIES IN A COMPLEX VARIABLEThis series is absolutely convergent if∞∑|a n |r n , (24.12)n=0which is a series of positive real terms, is convergent. Thus tests for the absoluteconvergence of real series can be used in the present context, and of these themost appropriate form is based on the Cauchy root test. With the radius ofconvergence R defined by1R = lim |a n| 1/n , (24.13)n→∞the series (24.10) is absolutely convergent if |z| R.If |z| = R then no particular conclusion may be drawn, and this case must beconsidered separately, as discussed in subsection 4.5.1.A circle of radius R centred on the origin is called the circle of convergenceof the series ∑ a n z n .ThecasesR = 0 and R = ∞ correspond, respectively, toconvergence at the origin only and convergence everywhere. For R finite theconvergence occurs in a restricted part of the z-plane (the Argand diagram). Fora power series about a general point z 0 , the circle of convergence is, of course,centred on that point.◮Find the parts of the z-plane for which the following series are convergent:∞∑ z n∞(i)n! , (ii) ∑∞∑n!z n z n, (iii)n .n=0n=0n=1(i) Since (n!) 1/n behaves like n as n →∞we find lim(1/n!) 1/n =0.HenceR = ∞ and theseries is convergent for all z. (ii) Correspondingly, lim(n!) 1/n = ∞. Thus R =0andtheseries converges only at z = 0. (iii) As n →∞,(n) 1/n has a lower limit of 1 and hencelim(1/n) 1/n =1/1 = 1. Thus the series is absolutely convergent if the condition |z| < 1issatisfied. ◭Case (iii) in the above example provides a good illustration of the fact thaton its circle of convergence a power series may or may not converge. For thisparticular series, the circle of convergence is |z| = 1, so let us consider theconvergence of the series at two different points on this circle. Taking z =1,theseries becomes∞∑ 1n =1+1 2 + 1 3 + 1 4 + ··· ,n=1which is easily shown to diverge (by, for example, grouping terms, as discussed insubsection 4.3.2). Taking z = −1, however, the series is given by∞∑ (−1) n= −1+ 1 n2 − 1 3 + 1 4 − ··· ,n=1831