# Showing a (complex) series is (conditionally) convergent.

1. Sep 20, 2010

### shoescreen

I've been reading a complex analysis book which had an example showing $$\sum^\infty_{n=1}1/n \cdot z^n$$ is convergent in the open unit ball.

I'm now looking at the case when $$|z| = 1$$. Clearly $$z = 1$$ is the divergent harmonic series, but i know this series is in fact convergent for all other $$|z| = 1$$.

In order to prove this, i need to be able to show that the related series $$\sum^\infty_{n=1}z^n$$ is bounded, whenever $$|z| = 1$$ and $$z \neq 1$$,.

I can solve this problem when ever the argument of z is a rational multiple of pi, but other than that I'm stuck. Any help proving that this related series is bounded would be very helpful.

Thanks!

2. Sep 20, 2010

### mathman

If you sum the truncated (at N) series, you have (1-zN+1)/(1-z). For z=1, you have 0/0 (no good). For z≠1, the expression is well defined, so see what happens as N -> ∞.