TITLE: Expressing $\sum_{n=-\infty}^\infty\dfrac{1}{z^3-n^3}$ in closed form
QUESTION [7 upvotes]: I want to express $$\sum_{n=-\infty}^\infty\dfrac{1}{z^3-n^3}$$ in closed form.
I know that $$\pi z\cot(\pi z)=1+2z^2\sum_{n=1}^\infty\dfrac{1}{z^2-n^2}$$ which looks close, but I don’t know how to use it.

REPLY [4 votes]: Assuming you are referring to the book by Ahlfors, here is a method relying on it:
Starting with a partial fraction decomposition w.r.t. $n$ we find
$$\frac{1}{z^3-n^3}=\frac{ \rho_3^0}{3z^2} \frac{1}{ \rho_3^0 z-n}+\frac{\rho_3^1}{3z^2} \frac{1}{ \rho_3 z-n}+\frac{\rho_3^2}{3z^2} \frac{1}{ \rho_3^2 z-n} $$ where $\rho_3=\exp(2 \pi i/3)$. Using the formula $$ \pi \cot( \pi z)= \lim_{m \to \infty} \sum_{n=-m}^m \frac{1}{z-n} $$
from page 189 three times should give you the desired result. I will add that since the series converges in the usual sense, it also converges w.r.t. this "principal value summation".