Behavior of $\sum_{n=1}^\infty n^{-1}z^n$ on the circle of convergence

Consider the following complex power series :$$\sum_{n=1}^\infty\frac{z^n}{n}$$ The radius of convergence of this series is $1$ and the series is divergent for $z=1$. I want to know what are the values of $z\in C:=\lbrace z\in\mathbb{C}: |z|=1\rbrace$, the circle of convergence, for which the given series converges.

• To find any value you may rewrite it as $\ -\ln(1-z)\$ (for $z\not =1$) – Raymond Manzoni Sep 10 '12 at 11:41

HINT: Look at Dirichlet test. In your case, choose $a_n = \dfrac1n$ and $b_n = z^n$.
From the Dirichlet test, you will get that the series converges everywhere on the boundary of the unit disc except at $z=1$.
• Although it ultimately amounts to the same, I'd suggest to consider $(1-z)f(z)$ where $f(z)$ is the given series. – t.b. Sep 10 '12 at 10:20
• According to this statement of Abel's Test, it doesn't apply, since $\sum\limits_{n=1}^\infty z^n$ doesn't converge. Would you explain what you mean by Abel's Test? – robjohn Sep 14 '12 at 6:03