# The infinite sum $\sum_{n=1}^{\infty}\frac{n^a}{e^{2\pi n}-1}$

I came across in a video that says

$$\sum_{n=1}^{\infty}\frac{3n^5(6596n^8+924n^4-2856)}{e^{2\pi n}-1}=818,$$

and that makes me wonder how on earth the infinite sum

$$\sum_{n=1}^{\infty}\frac{n^a}{e^{2\pi n}-1},$$

where $$a$$ is a positive integer, can sometimes have such a nice closed form or even be rational.

I plugged some polynomials into WolframAlpha, and in a split second I found

\begin{align*} \sum_{n=1}^{\infty}\frac{n}{e^{2\pi n}-1}&=\frac1{24}-\frac1{8\pi}\\ \sum_{n=1}^{\infty}\frac{n^5}{e^{2\pi n}-1}&=\frac1{504}\\ \sum_{n=1}^{\infty}\frac{n^9}{e^{2\pi n}-1}&=\frac1{264}\\ \sum_{n=1}^{\infty}\frac{n^{13}}{e^{2\pi n}-1}&=\frac1{24}, \end{align*}

the latter 3 gives

$$\sum_{n=1}^{\infty}\frac{3n^5(6596n^8+924n^4-2856)}{e^{2\pi n}-1}=\frac{19788}{24}+\frac{2772}{264}-\frac{8568}{504}=818.$$

I tried to calculate some results using residue theorem, where

$$f(z)=\frac{\pi z^a\cot(\pi z)}{e^{2\pi z}-1}.$$

This gives

\begin{align*} \operatorname{Res}(f,n)&=\frac{n^a}{e^{2\pi n}-1}\\ \operatorname{Res}(f,0)&=\begin{cases} \frac1{2\pi}&\text{if }a=1\\ 0&\text{if }a>1, \end{cases} \end{align*}

which seems to be what I want, but the integral

$$\oint_C\frac{\pi z^a\cot(\pi z)}{e^{2\pi z}-1}dz$$

stumbles me. After all, I want to avoid the poles at negative integers because the sum of $$\operatorname{Res}(f,n)$$ through negative integers diverges.

Could someone gives an insight of how to calculate the infinite sum to get a closed form for those having one?

### Edit

Two comments gave the answer I need, but I would like to know whether there are more elementary derivations of the formulas, say, only using elementary algebra and at the very most, residue theorem?

• Aug 25, 2022 at 5:29
• Aug 25, 2022 at 6:00
• @RobertZ Thanks, I can see from that it explains the $\frac1{8\pi}$ for $a=1$ now Aug 25, 2022 at 6:13
• @ClaudeLeibovici Thanks, it was indeed the exact image I saw! Aug 25, 2022 at 6:14