Find the limit $\lim_{z\to2k\pi i}\frac{z}{e^z-1}$ where $k\in\Bbb{Z}$

Question

If $k=0$, then ${e^z-1\over z}=\sum_{n\ge0}{z^n\over (n+1)!}$. So. $\lim_{z\to0}\frac{z}{e^z-1}=1$.
Now if $k\ne0$, then $\lim_{z\to2k\pi i}{e^z-1\over z}=\sum_{n\ge0}{(2k\pi i)^n\over (n+1)!}$, now can it be zero? If this limit is non-zero then the required limit will be a finite complex number, if not then the answer will be $\infty$.
So, my question is- Can $\sum_{n\ge0}{(2k\pi i)^n\over (n+1)!}$ be $0$?
Thanks for assistance in advance.

Answer 1

Since, if $k\neq0$, $\lim_{z\to2k\pi i}z=2k\pi i\neq0$ and $\lim_{z\to2k\pi i}e^z-1=0$,$$\lim_{z\to2k\pi i}\frac z{e^z-1}=\infty.$$