On proving that $\sum\limits_{n=1}^\infty \frac{n^{13}}{e^{2\pi n}-1}=\frac 1{24}$ 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.
 A: Theorem 1. (see [1] pg.275-276)
Let $a,b>0$ with $ab=\pi^2$, and let $\nu$ be any non zero integer. Then
$$
a^{-\nu}\left\{\frac{1}{2}\zeta(2\nu+1)+\sum^{\infty}_{n=1}\frac{n^{-2\nu-1}}{e^{2an}-1}\right\}-
(-b)^{-\nu}\left\{\frac{1}{2}\zeta(2\nu+1)+\sum^{\infty}_{n=1}\frac{n^{-2\nu-1}}{e^{2bn}-1}\right\}=
$$
\begin{equation}
=-2^{2\nu}\sum^{\nu+1}_{n=0}(-1)^n\frac{B_{2n}}{(2n)!}\frac{B_{2\nu+2-2n}}{(2\nu+2-2n)!}a^{\nu+1-n}b^n,\tag 1
\end{equation}
where $\zeta(s)$ is the Riemann zeta function and $B_n$ is the $n-$th Bernoulli number.
Notes
For integer $\nu<-1$ formula (1) evaluated from:
Theorem 2. (see [1] pg.261)
If $\nu$ is integer greater than 1, then ($ab=\pi^2$, $a,b>0$)
$$
a^{\nu}\sum^{\infty}_{n=1}\frac{n^{2\nu-1}}{e^{2an}-1}-(-b)^{\nu}\sum^{\infty}_{n=1}\frac{n^{2\nu-1}}{e^{2bn}-1}=(a^{\nu}-(-b)^{\nu})\frac{B_{2\nu}}{4\nu}\tag 2
$$
[1]: B.C. Berndt, 'Ramanujan`s Notebooks Part II'. Springer Verlang, New York., (1989).
A: 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.
A: This is not a strict solution but a heuristic proof using CAS. It also shows that there are other "magic" numbers instead of 13 which result in similar simple fractions. Playing around I also found the interesting conincidence that in some cases the sum and the related integral give the same results.
Defining
$$S(m) = \sum_{n=1}^{\infty} \frac{n^m}{e^{2 \pi n}-1}\tag{1}$$
we have to show that $S(13) = \frac{1}{24}$.
My first idea was to identify the denominator. I knew this expression from the Planck radioation formula and from the generating function of the Benoulli numbers.
But why so complicated? It is just the sum of a power series. Indeed we can write
$$\frac{1}{e^{2 \pi  n}-1} = \sum _{j=1}^{\infty } \exp (- 2 \pi j n)\tag{2}$$
Next, for the numerator we replace the power of $n$ by $z^n$ and consider the intermediate generating function
$$g_{0}(z,j) = \sum _{n=1}^{\infty } z^n \exp (- 2 \pi j n)=\frac{z}{e^{2 \pi  j}-z}\tag{3}$$
Summing over $j$ gives the generating function
$$\begin{align}g(z) = & \sum _{j=1}^{\infty } \frac{z}{e^{2 \pi  j}-z}\\=& -\frac{1}{2 \pi}\psi _{e^{-2 \pi }}^{(0)}\left(1-\frac{\log (z)}{2 \pi }\right)+\frac{1}{2 \pi} \log \left(1-e^{-2 \pi }\right)
\end{align}\tag{4}$$
Here $\psi _{q }^{(0)}(x)$ is the q-digamma function.
The the sums in question here can be found as derivatives of $g(z)$
$$S(m,z) = (z\frac{\partial}{\partial z})^m g(z) |_{z \to 1}\tag{5}$$
Using Mathematica for some values of $m$ I saw the pattern and found the general formula
$$S(m,z)=\frac{(-1)^{m+1} \psi _{e^{-2 \pi }}^{(m)}\left(1-\frac{\log (z)}{2 \pi }\right)}{2^{m+1} \pi ^{m+1}}\tag{6}$$
which gives at $z=1$ where the $\log$ vanishes the result
$$S(m)=S(m,z\to 1) = \frac{1}{(2 \pi )^{n+1}} \psi _{e^{-2 \pi }}^{(n)}(1)\tag{7}$$
Finally, the numerical results in Mathematica for $\frac{1}{S(m)}$ were particularly simple for 3 values of $m$. In the format {m,1/S(m)} we have
$$\{\{5, 504\}, \{9, 264\}, \{13,24\}\}\tag{8}$$
The case $m=13$ leads to the desired result. The "magic" values of $m$ have the form $1+4k$.
Discussion
It is always tempting with sums to look at the corresponding integral.
In our case we consider
$$i(m) = \int_0^{\infty } \frac{n^m}{\exp (2 \pi  n)-1} \, dn\tag{9}$$
where in contrast to the sum the integration starts at $n=0$.
The surprising observation is that we have (checked numerically)
$$i(m) = S(m),m=1+4k, k=1,2,...\tag{10}$$
which includes the "magic" values.
It would be nice to find a proof of this observation
A: It is the weight $14$ Eisenstein series
$$G_{14}(z)=\sum_{(n,m)\ne (0,0)} \frac1{(zn+m)^{14}}= 2\zeta(14)+\sum_{n\ne 0} \frac{1}{13!} \frac{d^{13}}{dz^{13}}\frac{2i\pi}{e^{2i\pi n z}-1}$$
$$=2\zeta(14)+\sum_{n\ge 1} \frac{4i\pi}{13!} \sum_{m\ge 1} (2i\pi m)^{13}e^{2i\pi mz}=2\zeta(14)+(2i\pi)^{14}\frac{2}{13!}\sum_{k\ne 1}\frac{k^{13}}{e^{-2i\pi kz}-1} $$
$$G_{14}(z)= z^{-14}G_{14}(-1/z)\implies \qquad G_{14}(i)=0$$
$$\boxed{(2i\pi)^{14}\frac{2}{13!}\sum_{k\ne 1}\frac{k^{13}}{e^{2\pi kz}-1}+2\zeta(14)=0 }$$
$2\zeta(14)=-\frac{B_{14}(2\pi)^{14}}{(14)!} $
A: For your curiosity !
I do not know if these results are known but, beside this one,
$$ \sum_{n=1}^\infty \frac{n^{5}}{e^{2\pi n}-1}=\frac 1{504}=\frac 1{21 \times 24}\qquad\text{and}  \qquad \sum_{n=1}^\infty \frac{n^{9}}{e^{2\pi n}-1}=\frac 1{264}=\frac 1{11 \times 24}$$
If they are known, please tell me where I could find them.
