closed form of $\sum \frac{1}{z^3 - n^3}$

Note that $$ \begin{align} \frac{3z^2}{z^3-n^3} &=\frac{1}{z-n}+\frac{1}{z-n\alpha}+\frac{1}{z-n/\alpha}\\ &=\frac{1}{z-n}+\frac{1/\alpha}{z/\alpha-n}+\frac{\alpha}{z\alpha-n}\tag{1} \end{align} $$ where $\alpha+1/\alpha=-1$.

It is fairly well known that $$ \sum_{n\in\mathbb{Z}}\frac{1}{z-n} = \pi\cot(\pi z)\tag{2} $$ where the principal value of the series in $(2)$ is intended: $\frac{1}{z}+\sum_{n=1}^\infty\left(\frac{1}{z-n}+\frac{1}{z+n}\right)$.

Combining $(1)$ and $(2)$, we get that $$ \sum_{n\in\mathbb{Z}}\frac{1}{z^3-n^3}=\frac{1}{3z^2}\left(\pi\cot(\pi z)+\tfrac{\pi}{\alpha}\cot(\tfrac{\pi}{\alpha}z)+\pi\alpha\cot(\pi\alpha z)\right) $$

I reopened this question by request, and I've accepted robjohn's answer. But I wanted to add that there is something a bit nontrivial (I think) at one step of his answer.

So we had:

$\displaystyle \sum \dfrac{1}{z^3 - n^3} = \sum \dfrac{-1}{z^2(n - z)} + \dfrac{-\omega}{z^2(n - \omega z)} + \dfrac{-\omega^2}{z^2 (n - \omega^2 z)}$

$= \displaystyle \frac{1}{z^2} \sum \dfrac{-1}{(n - z)} + \dfrac{-\omega}{(n - \omega z)} + \dfrac{-\omega^2}{(n - \omega^2 z)}$ and, as he mentioned, it is well known that $\displaystyle \sum \frac{1}{z-n} = \pi \cot \pi z$.

But the problem is that the individual sums $\displaystyle \sum \frac{1}{z-n}, \sum \frac{1}{\omega z-n}, \text{and} \sum \frac{1}{\omega ^2 z-n}$ diverge, so we are unable to rearrange the terms to form the three cotangent series. But we note that $\displaystyle \frac{1}{z-n} + \frac{\omega}{\omega z-n} +\frac{\omega^2}{\omega ^2 z-n}$ is the same as $\displaystyle \left(\frac{1}{z-n} + \frac{1}{n} \right)+ \left(\frac{\omega}{\omega z-n} + \frac{\omega}{n} \right)+\left( \frac{\omega^2}{\omega ^2 z-n} + \frac{\omega^2}{n}\right)$ as $\dfrac{1 + \omega + \omega^2}{n} \equiv 0$.

And each of the three individual sums $\displaystyle \sum \frac{1}{z-n} + \frac{1}{n}, \sum \frac{\omega}{\omega z-n} + \frac{\omega}{n}, \text{and} \sum \frac{\omega^2}{\omega ^2 z-n} + \frac{\omega^2}{n}$ do converge, absolutely even, if not evaluated at a pole. Within each series, now that we have absolute convergence, we can pair up the $n^{\text{th}}$ term with the $-n^{\text{th}}$ term to see that the subtracted constants add nothing to the overall series.

And that's why we can rearrange the series into three cotangents and a factor, as in robjohn's answer.

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.