<s>[INST] Expressing $\sum_{n=-\infty}^\infty\dfrac{1}{z^3-n^3}$ in closed form 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.
 [/INST] First rewrite it as:
$$\frac{1}{z^3} + \sum_{n=1}^\infty \left(\frac{1}{z^3-n^3} + \frac{1}{z^3+n^3}\right) = \frac{1}{z^3} +\sum_1^\infty \frac{2z^3}{z^6-n^6}$$
so we only need to find:
$$\sum_{n=1}^\infty \frac{1}{z^6-n^6}$$
Now, if $\omega$ is a primitive cube root of unity, there are functions $a(u),b(u),c(u)$ such that for real $u\neq v$:
$$\frac{1}{u^3-v^3} = \frac{a(u)}{u-v} +\frac{b(u)}{\omega u -v} + \frac{c(u)}{\omega^2u-v}$$
Setting $u=z^2$ and $v=n^2$, and finding a solution $\zeta$ so that $\zeta^2=\omega$, then we have:
$$\frac{1}{z^6-n^6} = \frac{a(z^2)}{z^2 - n^2}+\frac{b\left((\zeta z)^2\right)}{(\zeta z)^2-n^2} + \frac{c\left((\zeta^2 z)^2\right)}{(\zeta^2 z)^2 -n^2}$$
So we can use your first result to find:
$$\sum_{n=1}^\infty \frac{1}{z^6-n^6} = a(z^2)f(z) + b(\zeta^2z^2)f(\zeta z) + c(\zeta^4 z^2)f(\zeta^2 z)$$
where $$f(z)=\sum_{n=1}^\infty \frac{1}{z^2-n^2} = \frac{\pi z\cot(\pi z)-1}{2z^2}$$
That seems a bit messy, and requires jumping into complex numbers...
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