Prove that $\sum_{k=1}^{\infty} \large\frac{k}{\text{e}^{2\pi k}-1}=\frac{1}{24}-\frac{1}{8\pi}$ Prove that 
$$\sum_{k=1}^{\infty} \frac{k}{\text{e}^{2\pi k}-1}=\frac{1}{24}-\frac{1}{8\pi}$$
 A: Consider the function complex valued function

$$f(z)=\frac{\cot(\pi z)}{\sinh^2(z \pi)}$$

and integrate it along a quadratic contour $\mathcal{C}$ with verticies $\{( N/2,i N/2),(- N/2,i N/2),(- N/2,-i N/2),( N/2,-i N/2)\}$ where  $N$ is a (big) odd natural number. Note that the choice of such of kind verticies is necessary to bypass the poles  of $f(z)$. 
The result is (we travel counter-clockwise)
$$
\oint_{\mathcal{C}}f(z)dz=\int_{N/2}^{-N/2}dxf(x+i N/2)+i\int_{N/2}^{-N/2}dyf(i y+ N/2)+\\\int_{-N/2}^{N/2}dxf(x-i N/2)+i\int_{-N/2}^{N/2}dyf(i y+ N/2)
$$
we now have (this is an exercise for the reader) 
$\lim_{N\rightarrow\infty}f(x\pm i N/2)=\frac{\mp i}{\cosh^2(\pi x)}$ and that $\lim_{N\rightarrow\infty}f(i y\pm N/2)=0$. 
Furthermore it is elementary to show (the integrand is the derivative of $\tanh$) that $\int_{\mathbb{R}}\frac{1}{\cosh^2(\pi x)}dx=\frac{2}{\pi}$
which means that 
$$
\oint_{\mathcal{C}}f(z)dz=-2\frac{2i}{\pi}+2\cdot0=-\frac{4i}{\pi} \quad (\spadesuit)
$$
in the limit $N\rightarrow \infty$.
On the other hand we know by the residue theorem that (note that $(N-1)/2$ is an integer by construction)
$$
\oint_{\mathcal{C}}f(z)dz=2 \pi i\sum_{-(N-1)/2\leq k\leq (N-1)/2}\text{Res}(f(z),z=k)+\text{Res}(f(z),z=i k)
$$
it is now straightforward to show that
$$
\text{Res}(f(z),z=k)=\text{Res}(f(z),z=i k) = \frac{1}{\pi\sinh^2(\pi k)} \quad \text{if}\,\,k\neq0
$$
$$
\text{Res}(f(z),z=0) =-\frac{2}{3 \pi}
$$
which allows us to write that
$$
\oint_{\mathcal{C}} f(z)dz = 8 i\sum_{k=1}^{\infty}\frac{1}{\sinh^2(\pi k)}-\frac{4 i}{3} \quad (\heartsuit)
$$
In the limit of $N\rightarrow\infty$.
Now putting togehter $(\heartsuit)$ and $(\spadesuit)$ it follows that

$$
\sum_{k=1}^{\infty}\frac{1}{\sinh^2(\pi k)}=\frac{1}{6}-\frac{1}{2\pi} \quad\blacksquare
$$

this proves the result used by @Fabian

Appendix
if we set $f_a(z)=\frac{\cot(\pi z)}{\sinh^2(a z \pi)}$ and follow the same steps as above we can show that on the one hand the sum of residues equals 
$$
\oint_{\mathcal{C}} f_a(z)dz=4i\sum_{k\geq1}\frac{1}{\sinh^2(k \pi a)}+\frac{4i}{a^2}\sum_{k\geq1}\frac{1}{\sinh^2(k \pi/ a)}-\frac {2 i}{3}\frac{1+a^2}{a^2}\quad (\heartsuit \heartsuit)
$$
and on the other hand, by parametrizing the contour of integration as shown above, we get that
$$
\oint_{\mathcal{C}} f_a(z)dz=-\frac{4i}{\pi a} \quad (\spadesuit \spadesuit)
$$
Now putting togehter $(\heartsuit\heartsuit)$ and $(\spadesuit\spadesuit)$ it follows that

$$
\sum_{k\geq1}\frac{1}{\sinh^2(k \pi a)}+\frac{1}{a^2}\sum_{k\geq1}\frac{1}{\sinh^2(k \pi/ a)}=\frac{1+a^2}{6 a^2}-\frac{1}{\pi a}\quad\blacksquare\quad\blacksquare
$$

in accordance with Berndt et al.
A: If we define Dedekind eta function $\eta(q)$ via the equation $$\eta(q) = q^{1/12}\prod_{i = 1}^{\infty}(1 - q^{2i})\tag{1}$$ then it can be proved via functional equation of Dedekind eta function that $$n^{1/4}\eta(e^{-\pi\sqrt{n}}) = \eta(e^{-\pi/\sqrt{n}})\tag{2}$$ Let's define function $P(q)$ as $$P(q) = 1 - 24\sum_{i = 1}^{\infty}\frac{iq^{2i}}{1 - q^{2i}} = 12q\frac{d}{dq}\log\eta(q)\tag{3}$$ Logarithmic differentiation of equation $(2)$ with respect to $n$ gives us $$nP(e^{-\pi\sqrt{n}}) + P(e^{-\pi/\sqrt{n}}) = \frac{6\sqrt{n}}{\pi}\tag{4}$$ and putting $n = 1$ we get $$P(e^{-\pi}) = \frac{3}{\pi}\tag{5}$$ and looking at the definition of $P(q)$ we see that the above equation is exactly what is asked in the question.
A: Rewrite
$$\frac{1}{e^{2\pi k} -1} = \sum_{n=1}^\infty e^{-2\pi k n}.$$
So we need to evaluate
$$\sum_{n,k=1}^\infty k e^{-2\pi k n}.$$
Summing first over $k$, we have
$$ \sum_{k=1}^\infty k e^{-2\pi k n} = \frac1{2\pi}\frac{\partial}{\partial n} \sum_{k=1}^\infty e^{-2\pi k n} = \frac1{2\pi}\frac{\partial}{\partial n} \frac{1}{e^{2\pi n} -1} =\frac{e^{2\pi n}}{(e^{2\pi n}-1)^2} = \frac{1}{4 \sinh^2(\pi n)} .$$
The sum
$$\sum_{n=1}^\infty \frac{1}{\sinh^2(\pi n)} =\frac{1}{6} - \frac{1}{2\pi} $$ is evaluated here, see also page 3 here, and the quoted result follows.
