TITLE: On proving that $\sum\limits_{n=1}^\infty \frac{n^{13}}{e^{2\pi n}-1}=\frac 1{24}$
QUESTION [23 upvotes]: Ramanujan found the following formula:

$$\large \sum_{n=1}^\infty \frac{n^{13}}{e^{2\pi n}-1}=\frac 1{24}$$

I let $e^{2\pi n}-1=\left(e^{\pi n}+1\right)\left(e^{\pi n}-1\right)$ to try partial fraction decomposition and turn the sum into telescoping, but methinks it doesn't lead anywhere and only makes things hairy.
How does one go about proving this? Thanks.

REPLY [3 votes]: Essentially the same as @reuns answer but with more detail.
Let $k\ge2$ be an integer and define the Eisenstein series
$$G_{2k}(z)=\sum_{(n,m)\in A}\frac{1}{(n+mz)^{2k}},\tag 1$$
where $A=\Bbb Z^2\setminus\{(0,0)\}$, and $z\in\Bbb C$ with $\Im(z)>0$. It is simple to show that $G_{2k}(z+1)=G_{2k}(z)$ for all $z$, so it follows that we may write
$$G_{2k}(z)=\sum_{n\ge0}g_nq^n,$$
where $q=e^{2i\pi z}$. It can be shown (see here) that there is a closed form for $g_n$. Namely, we can write
$$\begin{align}
G_{2k}(z)&=2\zeta(2k)\left(1+c_{2k}\sum_{n\ge1}\sigma_{2k-1}(n)q^n\right)\\
&=2\zeta(2k)\left(1+c_{2k}\sum_{n\ge1}\frac{n^{2k-1}q^n}{1-q^n}\right),
\end{align}$$
where $c_{2k}=\frac{(2\pi i)^{2k}}{(2k-1)!\zeta(2k)}=\frac{-4k}{B_{2k}}=\frac{2}{\zeta(1-2k)}$.
On the other hand, it is simple to show from $(1)$ that $$G_{2k}(-1/z)=z^{2k}G_{2k}(z).$$
Letting $E_{2k}(z)=\frac{1}{2\zeta(2k)}G_{2k}(z)$ for convenience, we have
$$E_{2k}(-1/z)=z^{2k}E_{2k}(z).\tag2$$
Then defining $$S_k(q)=\sum_{n\ge1}\frac{n^{2k-1}}{q^n-1},$$
we have $$E_{2k}(z)=1+c_{2k}S_k(e^{-2i\pi z}).$$
Then from $(2)$, we have
$$1+c_{2k}S_k(e^{2i\pi/z})=z^{2k}\left(1+c_{2k}S_k(e^{-2i\pi z})\right).\tag3$$
Since your sum is given by $S_7(e^{2\pi})$, we set $k=7$ and $z=i$ in $(3)$, and get
$$\begin{align}
1+c_{14}S_7(e^{2\pi})&=-\left(1+c_{14}S_7(e^{2\pi})\right)\\
\Rightarrow S_7(e^{2\pi})&=-\frac{1}{c_{14}}.
\end{align}$$
Since $c_{2k}=-4k/B_{2k}$, we have $-1/c_{14}=B_{14}/28=1/24$, and thus
$$S_7(e^{2\pi})=\sum_{n\ge1}\frac{n^{13}}{e^{2\pi n}-1}=\frac{1}{24}.$$
In general, the same method allows us to compute
$$S_{2k+1}(e^{2\pi})=\sum_{n\ge1}\frac{n^{4k+1}}{e^{2\pi n}-1}=\frac{B_{4k+2}}{8k+4},$$ as was shown by @MarcoRiedel.