Determining precisely where $\sum_{n=1}^\infty\frac{z^n}{n}$ converges? Inspired by the exponential series, I'm curious about where exactly the series $\displaystyle\sum_{n=1}^\infty\frac{z^n}{n}$ for $z\in\mathbb{C}$ converges. 
I calculated
$$
\limsup_{n\to\infty}\sqrt[n]{\frac{1}{n}}=\limsup_{n\to\infty}\frac{1}{n^{1/n}}
$$
and
$$
\lim_{n\to\infty} n^{1/n}=e^{\lim_{n\to\infty}\log(n)/n}=e^{\lim_{n\to\infty}1/n}=e^0=1.
$$
So the radius of convergence is $1$, so the series converges on all $z$ inside $S^1$. But is there a way to tell for which $z$ on the unit circle the series converges? I know it converges for $z=-1$, but diverges for $z=1$, but I don't know about the rest of the circle. For what other $z$ does this series converge? Thanks. 
 A: The following theorem on power series is due to E. Picard:

Let $(a_n)$ be a sequence of real numbers.
If the sequence $(a_n)$ is nonnegative, decreases and tends to zero when $n\to \infty$, then the complex power series $\sum a_n\ z^n$ converges in the closed unit disc $\overline{D}(0;1)$ with the only possible exception of the point $1$.

The proof of Picard's theorem relies on Abel's summation by parts formula, as far as I remember.
Now, the coefficients of your series, i.e. $a_n=1/n$, satisfy the assumptions of Picard's theorem, hence your series converges at least in $\overline{D}(0;1)\setminus \{1\}$; on the other hand, the series diverges when $z=1$ (for it becomes the harmonic series).
Therefore the convergence set of $\sum 1/n\ z^n$ is $\overline{D}(0;1)\setminus \{1\}$.
A: Fix $z$ in the unit circle, i.e. $|z|=1$. We want to apply Dirichlet's test: if $\{a_n\}$ are real numbers and $\{b_n\}$ complex numbers such that:


*

*$a_1\geq a_2\geq\cdots$

*$\lim_{n\to\infty}a_n=0$

*There exists $M>0$ such that $\left|\sum_{n=1}^Nb_n\right|\leq M$ for every $N\in\mathbb{N}$;


then $\sum_{n=1}^\infty a_nb_n$ converges. Here $a_n=1/n$, $b_n=z^n$. The first two conditions are clearly satisfied, and for the third one:
$$
\left|\sum_{n=1}^Nz^{ n }\right|=\left|\frac{z-z^{N+1}}{1-z}\right|\leq\frac{2}{|1-z|}
$$
for all $N\in\mathbb{N}$. 
This shows that the third condition is satisfied for every $z\ne1$ in the circle. 
In conclusion, the series converges for every $z$ with $|z|\leq1$ other than $z=1$, and it diverges for $|z|>1$.
A: Even though the 1st answer is the more easy and appropriate here is a more analytic one. $\sum_{n=1}^{k} \frac{z^n}{n} =\frac{z}{1-z} $$\sum_{n=1}^{k-1} [\frac{1-z^n}{n(n+1)} + \frac{1-z^k}{k}]$
(we can prove that with induction till k).
We can also see that
$0 \leqslant | \frac{1-z^k}{k}|\leqslant \frac{2}{k}\rightarrow 0$ for k$\rightarrow$$\infty$
also,
$0 \leqslant |\frac{1-z^n}{n(n+1)}|\leqslant \frac{2}{n(n+1)} $
$\sum_{n=1}^{\infty}\frac{2}{n(n+1)}$ converges so
$\sum_{n=1}^{\infty} \frac{z^n}{n}$ converges as well
