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

[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!

 

[mathman]

If you sum the truncated (at N) series, you have (1-z^N+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 -> ∞.