TITLE: closed form of $\sum \frac{1}{z^3 - n^3}$
QUESTION [17 upvotes]: I am currently trying to find a closed form expression for $\displaystyle f(z) = \sum_{n \in \mathbb{Z}} \frac{1}{z^3 - n^3}$, $z \in \mathbb{C}$. After a set of twists and turns, I have hit a wall.
I currently believe the best way to proceed is to use a partial fractions. So when we see that $\dfrac{1}{z^3 - n^3} = \dfrac{1}{n^2(z - n)} + \dfrac{\omega}{n^2(z - \omega n)} + \dfrac{\omega^2}{n^2(z - \omega^2 n)} = $
$= \dfrac{-1}{z^2(n - z)} + \dfrac{-\omega}{z^2(n - \omega z)} + \dfrac{-\omega^2}{z^2 (n - \omega^2 z)}$ (depending on which way you want to factor, and where $\omega$ is the first cubic root unity.
My overall thought process is to, more or less, try to come up with a function whose poles and residues at those poles agree - then their difference is an entire function. And if we're witty, the difference may even be zero. In these partial fractions forms, it is very easy to see what the poles and residues of my function are. But I have not been able to find a convenient function with those poles (that will lead to some closed form at the end of the day).
This is actually an exercise from Ahlfors, pg 191 number 2. The section it is in is dedicated to Mittag-Leffler, which is part of the reason why I am so predisposed to using a root comparison strategy.
One would assume that if we are able to do it for any of the three terms of the partial fraction decomposition, then we are set. I will note, however, that the first of my two given deconpositions behaves much better than the second (as all the terms converge very nicely as long as we don't consider the poles in the first, but each series diverges on its own in the second).

REPLY [10 votes]: Here is another way.  I am posting it because I believe it provides some more insight into the problem.  What we are really doing is just extracting every 3rd coefficient from the cotangent series.  That is why the cube roots of unity appear.
Notice that $$\sum_{n\in\mathbb{Z}} \frac{1}{z^3-n^3}=\frac{1}{z^3}+\sum_{n=1}^\infty \frac{2z^3}{z^6-n^6}.$$  
Then this is 
$$\frac{1}{z^{3}}+2z^{3}\sum_{n=1}^{\infty}\frac{1}{z^{6}-n^{6}}=\frac{1}{z^{3}}-2z^{3}\sum_{n=1}^{\infty}\frac{1}{n^{6}}\frac{1}{1-\frac{z^{6}}{n^{6}}}=\frac{1}{z^{3}}-2z^{3}\sum_{n=1}^{\infty}\frac{1}{n^{6}}\sum_{m=0}^{\infty}\frac{z^{6m}}{n^{6m}} $$
$$=\frac{1}{z^{3}}-2z^{3}\sum_{m=0}^{\infty}z^{6m}\zeta(6+6m).$$ Now, as $$\frac{\pi\cot(\pi x)}{x^{2}}=\frac{1}{x^{3}}-2\sum_{n=0}^{\infty}x^{2n-1}\zeta(2n+2)$$ (same proof as before) we see that we are just extracting every 6th coefficients. To to this for general $f(z)$  we look at $f(z)+f(\zeta z)+f(\zeta^{2}z)$  where $\zeta$ is a cube root of unity.  In this case, a small computation yields $$\frac{\pi\cot(\pi x)}{x^{2}}+\frac{\pi\cot(\pi\zeta x)}{\zeta^{2}x^{2}}+\frac{\pi\cot(\pi\zeta^{2}x)}{\zeta x^{2}}=3\left(\frac{1}{x^{3}}-2\sum_{n=0}^{\infty}x^{6n+3}\zeta(6n+6)\right).$$ Thus we conclude that $$\sum_{n\in\mathbb{Z}}\frac{1}{z^{3}-n^{3}}=\frac{\pi\cot(\pi z)+\pi\zeta\cot(\pi\zeta x)+\pi\zeta^{2}\cot(\pi\zeta^{2}z)}{3z^{2}}. $$
Remark: There is also an answer by Qiaochu Yuan on the subject of extracting coefficients.