TITLE: singularity of analytic continuation of $f(z) = \sum_{n=1}^\infty \frac{z^n}{n^2}$
QUESTION [5 upvotes]: How to show that all possible collection of analytic continuations of $\displaystyle f(z) = \sum_{n=1}^\infty \frac{z^n}{n^2} $ has singular point at $z = 1$. I know that $f(z)$ converges for $|z| \le 1$. Also is there a theorem that relates the singularity of analytic continuation with circle of convergence?

REPLY [3 votes]: Note that when $|z|<1$,
$$f'(z)=\sum_{n=1}^\infty\frac{z^{n-1}}{n}=-\frac{\log(1-z)}{z}.\tag{1}$$
Since the right hand side of $(1)$ has a unique singularity at $z=1$, it implies that on the unit circle, the analytic continuation of $f$ has a unique singularity at $z=1$. 
For the general situation, note that for one complex variable, any (nonempty) open set in $\mathbb{C}$ is a domain of holomorphy, which implies that for any closed subset $S$ of the unit circle, there exists a holomorphic function $f$ defined on the unit disk, such that the collection of sigularities of the analytic continuation of $f$ on the unit circle is precisely $S$.  

Remark: Just in case, $(1)$ follows from integrating
$$(zf'(z))'=\sum_{n=0}^\infty z^n=\frac{1}{1-z}.$$