# Help Calculation of $\sum_{k=1}^{\infty} \frac{k^{2n}}{e^k -1}$

Recently, I read a book : Euler, Riemann, Ramanujan - Contact mathematician beyond the space-time by Nobushige Kurokaw. It says that Ramanujan had found the following formula

$$\sum_{k=1}^{\infty} \frac{k}{e^{2k \pi}-1}=\frac{1}{24}-\frac{1}{8\pi}$$

After few month, I succeeded in finding similar formula using Euler-Maclaurin Formula:

$$\sum_{k=1}^{\infty}\frac{k}{e^k -1}=\frac{{\pi}^2}{6}-\frac{11}{24}$$ $$\sum_{k=1}^{\infty}\frac{k^{2n-1}}{e^k -1}=\frac{\left| B_{2n}\right|}{4n}((2\pi)^{2n}+(-1)^{n+1}) \quad when \quad n>1$$

I wonder if we can generalize te following formula :

$$\sum_{k=1}^{\infty}\frac{k^{2n}}{e^k -1}$$

I tried with various ways, but I failed. Please answer back users~~~

PS. It's first time that I answer a question on this site. So I could have made some mistakes while writing....

• Do you have a proof for the second series $\sum k/(e^{k} - 1)$? I wonder how it got such a nice closed form. – Paramanand Singh Sep 1 '17 at 15:15
• $$\sum_{k=1}^{\infty}\frac{k}{e^k -1} \neq \frac{{\pi}^2}{6}-\frac{11}{24}$$ $$\text{RHS} = 1.18660073351489\color{red}{3103139081833312691}$$ $$\text{LHS} = 1.18660073351489\color{red}{2820585500128198498}$$ If you obtained it via Euler-Maclaurin Formula, you might have omitted remainder terms. You might want to experiment these on Mathematica yourself. :) – pisco Sep 1 '17 at 15:59
• Thanks @pisco125 for disproving the identity. The Ramanujan function $P(q)=1-24\sum_{k=1}^{\infty}\frac{kq^{k}}{1-q^{k}}$ (which is related to the series in question) has a known closed form when $q=\pm e^{-\pi\sqrt{n}}$ for positive rational $n$. – Paramanand Singh Sep 1 '17 at 16:09
• @ParamanandSingh I was a bit skeptical when I first saw this identity by OP. Because if it were true, then there would be no reason that Ramanujan did not notice it. I have seen the close form $e^{-\pi}$ and $e^{-\sqrt{2}\pi}$, are they related elliptic singular values? Does one of Ramanujan's notebook contain information about them? I am also very curious about these sums. – pisco Sep 1 '17 at 16:18
• @pisco125: yes they are related to elliptic singular values. The function $P(q)$ can be expressed in terms of elliptic integrals as $$P(q) =\left(\frac{2K}{\pi}\right)^{2}\left(\frac{6E}{K}+k^{2}-5\right)$$ – Paramanand Singh Sep 1 '17 at 18:21

CAS says:

$$\sum _{k=1}^{\infty } \frac{k^{2 n}}{\exp (k)-1}=\sum _{j=1}^{\infty } \left(\sum _{k=1}^{\infty } k^{2 n} \exp (-j k)\right)=\sum _{j=1}^{\infty } \text{Li}_{-2 n}\left(e^{-j}\right)=-\psi _e^{(2 n)}(1)$$

where:

$\text{Li}_{-2 n}\left(e^{-j}\right)$ is the polylogarithm function,

$\psi _e^{(2 n)}(1)$ is the n-th derivative of the q-digamma function.

Mathematica code:

Sum[k^(2 n)/(Exp[k] - 1), {k, 1, Infinity}] == -QPolyGamma[2 n, 1, E]

where E is the exponential constant.