Convergence of a Taylor Series

[tylerc1991]

Homework Statement

Suppose that: sum [a_n (n-1)^n] is the Talyor series representation of tanh(z) at the point z = 1. What is the largest subset of the complex plane such that this series converges?

Note: 'sum' represents the sum from n=0 to infinity

Homework Equations

tanh(z) = sinh(z)/cosh(z) ; cosh(z) = 0 for z = pi*i/2

B(C,R) is the open ball centered at C and with radius R

d(x,y) is the distance between x and y

The Attempt at a Solution

So the series is centered at z = 1 and the first place that tanh(z) is not analytic (expanding from z = 1) is at the point z = pi*i/2. so I can create an open ball of radius d(1,pi*i/2) and this is the largest subset of convergence.

d(1,pi*i/2) = (1/2)sqrt(4 + pi^2)

so the largest subset in which the Taylor series converges is B(1,(1/2)sqrt(4 + pi^2))

 

[Dick]

I believe that would be correct. The radius of convergence is the distance to the nearest singularity. Did you have a question?