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

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).

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(+1) for showing your work :) –  The Chaz 2.0 Nov 4 '11 at 1:14

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

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Ok I am nitpicking a bit here, but I think you should mention somewhere that none of the series you are working converge well enough to justify the above operations. Of course everything works out, and is correct, but I don't think $\sum_{n\in\mathbb{Z}}\frac{1}{z-n} = \pi\cot(\pi z)$ is true, let alone well known. We do have $\lim_{N\rightarrow \infty} \sum_{n=-N}^N\frac{1}{z-n} = \pi\cot(\pi z)$, but that is slightly different. –  Eric Naslund Nov 4 '11 at 1:53
You are correct. The series in $(2)$ does not converge absolutely. However, its principal value, $\frac{1}{z}+\sum_{n=1}^\infty\left(\frac{1}{z-n}+\frac{1}{z+n}\right)$ does. That is what I mean by $(2)$. I will mention this in my answer. –  robjohn Nov 4 '11 at 7:22

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.

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

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

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I guess a (+1) is in order here, too ;) –  The Chaz 2.0 Nov 4 '11 at 1:14

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.

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Thanks, pretty cool technique. –  Jack Schmidt Nov 4 '11 at 3:11