Prove that $$\sum_{k=1}^{\infty} \frac{k}{\text{e}^{2\pi k}-1}=\frac{1}{24}-\frac{1}{8\pi}$$
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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. |
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