Originally Posted by

**lpd** Can some one help me with this question?

Use the Dirichlet test to show that the series $\displaystyle \displaystyle\sum_{n=1}^{\infty}\frac{z^n}{n}

$.

converges for all $\displaystyle z$ with $\displaystyle |z|=1$ but $\displaystyle z \not=1$.

I know that the Dirichlet test for the convergence of a series of complex numbers can be stated as follows:

Suppose that $\displaystyle a_n$ is a sequence of positive real numbers with $\displaystyle a_n>a_{n+1}$ for all $\displaystyle n$ and $\displaystyle \displaystyle\lim_{n\to\infty}a_n$ = 0. Suppose that $\displaystyle b_n$ is a sequence of complex numbers so that there exists $\displaystyle M$ so that, for all $\displaystyle N$, $\displaystyle |\sum_{n=1}^{\infty}b_n |<M$. (Upper limit in that sum should be N, not infty.) Then the series $\displaystyle \sum_{n=1}^{\infty}a_n b_n$ converges.

So how do I start? Just really confused. Do I let $\displaystyle b_n= z^n$ and start from there? Yes !