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?
