Finding radius of convergence for taylor series of $\frac{z}{e^z-1}$ Im trying to find the radius of convergence for the taylor sum of the function $\frac{z}{e^z-1}$ around z=0.
So far I've found the coffiecients $a_0=1$, $a_1=-\frac{1}{2}$, $a_2=\frac{1}{12}$ and that $a_{2k+1}=0$ for $k\geq 1$. It's pretty ugly to look for the rest of the terms in the series.
Is there a neat way to find that? 

Hint: The radius of convergence of a series at $z_0$ is the distance from $z_0$ to (one of) its closest non-removable singularity. 
