Summing $\frac{1}{e^{2\pi}-1} + \frac{2}{e^{4\pi}-1} + \frac{3}{e^{6\pi}-1} + \cdots \text{ad inf}$ In this post, David Speyer, actually, gave an expression for $\displaystyle  \frac{t}{e^{t}-1}$.
The question is can we sum the given series, using that expression, if not how does one sum this series. $$\sum\limits_{n=1}^{\infty} \frac{n}{e^{2\pi n}-1}=\frac{1}{e^{2\pi}-1} + \frac{2}{e^{4\pi}-1} + \frac{3}{e^{6\pi}-1} + \cdots \text{ad inf}$$
 A: What you require here are the Eisenstein series. In particular the evaluation of
$$E_2(\tau) = 1 – 24\sum_{n=1}^\infty \frac{ne^{2\pi i n \tau} }{1 - e^{2\pi i n \tau}},$$
at $\tau = i. $ Rearrange to get
$$\sum_{n=1}^\infty \frac{ne^{2\pi i n \tau} }{1 - e^{2\pi i n \tau} } = \frac{1}{24}(1 – E_2(i) ).$$
See Lambert series for additional information.
EDIT: The function
$$G_ 2(\tau) = \zeta(2) \left(
1 – 24\sum_{n=1}^\infty \frac{ne^{2\pi i n \tau} }{1 - e^{2\pi i n \tau}} \right)
=\zeta(2)E_2(\tau)$$
satisfies the quasimodular transformation
$$G_ 2\left( \frac{a\tau+b}{c\tau+d} \right) =
(c\tau+d)^2G_ 2(\tau) - \pi i c (c\tau + d).$$
And so with $a=d=0,$ $c=1$ and $b=-1$ we find $G_ 2(i) = \pi/2.$ Therefore
$$E_2(i) = \frac{ G_ 2( i)}{ \zeta(2)} = \frac{\pi}{2}\frac{6}{\pi^2} = \frac{3}{\pi}.$$
Hence we obtain
$$\sum_{n=1}^\infty \frac{n}{e^{2\pi n} – 1} = \frac{1}{24} - \frac{1}{8\pi},$$
as given in the comment to the question by Slowsolver.
EDIT:
There is a very nice generalisation of the sum in the question.
For odd $ m > 1 $ we have
$$\sum_{n=1}^\infty \frac{n^{2m-1} }{ e^{2\pi n} -1 } = \frac{B_{2m}}{4m},$$
where $B_k$ are the Bernoulli numbers defined by
$$\frac{z}{e^z - 1} = \sum_{k=0}^\infty \frac{B_k}{k!} z^k \quad \textrm{ for }
|z| < 2 \pi.$$
