TITLE: Prove that $\sum_{k=1}^{\infty} \large\frac{k}{\text{e}^{2\pi k}-1}=\frac{1}{24}-\frac{1}{8\pi}$
QUESTION [45 upvotes]: Prove that 
$$\sum_{k=1}^{\infty} \frac{k}{\text{e}^{2\pi k}-1}=\frac{1}{24}-\frac{1}{8\pi}$$

REPLY [6 votes]: 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.