singularity of analytic continuation of $f(z) = \sum_{n=1}^\infty \frac{z^n}{n^2}$ 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?
 A: 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
$$(zf'(z))'=\sum_{n=0}^\infty z^n=\frac{1}{1-z}.$$
A: This is one of the polylogarithm, also $s=2$ case is called dilogarithm as GEdgar pointed out. 
The definition is 
$$
\operatorname{Li}_s(z) = \sum_{k=1}^\infty {z^k \over k^s} = z + {z^2 \over 2^s} + {z^3 \over 3^s} + \cdots \,
$$
Your case it is $s=2$, so $\operatorname{Li}_2(z)$. 
According to http://en.wikipedia.org/wiki/Polylogarithm, we have a integral representation of $\operatorname{Li}_s(z)$:
$$
\operatorname{Li}_{s}(z) = {1 \over \Gamma(s)}
\int_0^\infty {t^{s-1} \over e^t/z-1} \,\mathrm{d}t \,.
$$
So in your case 
$$
\operatorname{Li}_{2}(z) =  
\int_0^\infty {zt  \over e^t  - z} \,\mathrm{d}t \,.$$
This allows an analytic continuation to $z\in \mathbb{C}-[1,\infty)$. 
