# Radius of convergence of complex power series

1. Oct 6, 2009

1. The problem statement, all variables and given/known data
Find the radius of convergence of the series:

∑ n^-1.z^n
n=1

Use the following lemma:
∞ ∞
If |z_1 - w| < |z_2 - w| and if ∑a_n.(z_2 - w)^n converges, then ∑a_n.(z_1 - w)^n also
n=1 n=1
converges.
The contrapositive is also true.

2. Relevant equations

3. The attempt at a solution
Hi, here's what I've done:

Let r be the radius of convergence. The series converges when |z-w| < r and diverges for |z-w|> r.

In this question, z = z, w = 0.

Use the ratio test for convergence:

lim |a_n+1 / a_n| = lim |zn / n+1 |
n->∞
= |z|

Thus the series converges when |z| < 1
diverges when |z| > 1

So let |z - 0| = |z_2 - w|
Then, by the lemma, all |z_1 - 0| converges, where |z_1 - 0| < |z - 0|
and the converse is true for a diverging series.

Thus the radius of convergence = |z|

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I think I've covered everything, but I don't think I've made very satisfactory use of the lemma. Can anyone please point me in the right direction?
Thanks for any help.

2. Oct 7, 2009

### HallsofIvy

Staff Emeritus
Your final answer appears to be "Thus the radius of convergence is |z|" which is meaningless. the radius of convergence is a real number, not a variable.

You had already said "Thus the series converges when |z| < 1
diverges when |z| > 1". What more do you need?