Evaluate the series $ \sum_{n=1}^{\infty} \frac{1}{n(e^{2\pi n}-1)} $.
$$ \frac{1}{n(e^{2\pi n}-1)}=\sum_{k=0}^\infty \frac{e^{-2 \pi (k+1) n}}{n}$$ $$\sum_{n=1}^\infty \frac{e^{-2 \pi (k+1) n}}{n}=\log \left(1-e^{-2 \pi (k+1)}\right)$$ So, now, you need to compute $$\sum_{k=0}^\infty \log \left(1-e^{-2 \pi (k+1)}\right)$$ and this is explained in the link provided by @Jean-Marie.