Show convergence of $\sum \frac{z^n}{n}$ Show that the series $\displaystyle\sum \frac{z^n}{n}$ for $z=1$ diverges, but for all $z \ne 1$ with $|z|=1$ converges. Hint: Estimate $(1-z) \displaystyle\sum_{n=k}^{m} \frac{z^n}{n}$.
The first case $z=1$ is easy, this is just the harmonic series. But I am really stuck with the second part, which is the actually interesting one. First I tried to write $z=re^{i \phi}$, but the hint does not look like this could be useful. Specifically, I don't know how to bring the series in a form such that the hint is even useful.
Edit: To clearify some questions from the comments. The hint has no misprint and I don't see how the question about the uniform convergence of this series on the open unit disk that was posted multiple times in the comments helps me to solve this problem. However, thank you already for your help! There might be a good strategy by applying the Dirichlet test as one of the answers suggests.

Hint:
$$\sum_{k=1}^n\frac{z^k}{k}-\sum_{k=1}^n\frac{z^{k+1}}k=\sum_{k=1}^n\frac{z^k}{k}-\sum_{k=2}^{n+1}\frac{z^k}{k-1}=z-\sum_{k=2}^{n}\frac{z^k}{k(k-1)}-\frac{z^{n+1}}n.$$
