How to find all values of $z$ at which $\sum\limits_{n=1}^{\infty} \frac{1}{n^2} \exp(\frac{nz}{z-2})$ converges? Could anyone advise me on how to find all $z$ such that $\begin{align} \sum^{\infty}_{n=1} \dfrac{1}{n^2} \end{align}\text{exp}\left(\dfrac{nz}{z-2}\right)$ converges ? Does it suffice to find all $z$ such that $ \begin{align}\left|\text{exp}\left(\dfrac{nz}{z-2}\right)\right|\leq 1 \ ?\end{align}$
Hints will suffice, thank you. 

The radius of convergence of $\sum \frac{1}{n^2} x^n$ is $1$. Plug in $x = \exp(\frac{z}{z-2})$, so that the series converge iff
$$|\exp(\frac{z}{z-2})| < 1 \iff \Re\left( \frac{z}{z-2} \right) < 0.$$
By writing down $z = a+bi$, where $a,b \in \mathbb{R}$, you get (let's assume $z \neq 2$ all the time):
$$\begin{align}
\Re \frac{a+bi}{a-2+bi} < 0 & \iff \Re\frac{(a+bi)(a-2-bi)}{(a-2)^2+b^2} < 0 \\
& \iff \frac{a^2-2a+b^2}{(a-2)^2 + b^2} < 0 \\
& \iff a^2-2a+b^2 < 0.
\end{align}$$
So that's the region of convergence, with the caveat that $(a,b) \neq (2,0)$. (There remains the problem at the boundary)
