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|

---

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?