# 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. $$\sum\limits_{n=1}^{\infty} \frac{n}{e^{2\pi n}-1}=\frac{1}{e^{2\pi}-1} + \frac{2}{e^{4\pi}-1} + \frac{3}{e^{6\pi}-1} + \cdots \text{ad inf}$$

• Mathematica says that the above infinite sum is: $1/24−1/8\pi$. So, maybe a reasonably simple way to sum this series can be found. – user1709 Nov 4 '10 at 11:24
• This sum reappeared at this MSE link. – Marko Riedel Jan 29 '14 at 1:55

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

$$\frac{z}{e^z - 1} = \sum_{k=0}^\infty \frac{B_k}{k!} z^k \quad \textrm{ for } |z| < 2 \pi.$$

• @Derek: Super work! I don't know how did you mange to start with Eisenstein series directly. Is it experience or intuition. I couldn't even think of this, more over i have never studies Eisenstein series. The only place where i have even heard about eisenstein is his irreducibility criteria. – anonymous Nov 4 '10 at 16:30
• @Chandru1 It was experience actually. Eisenstein series are in my PhD thesis. I love anything Ramanujan related! – Derek Jennings Nov 4 '10 at 16:40
• @Derek: The only thing which i have managed to read from a Ramanujan's paper is : "On a question 330 of Prof. Sanjana" where he sums a given series, and i loved the idea. – anonymous Nov 4 '10 at 16:48
• @Derek: Thanks a lot sir, for the kind explanation! – anonymous Nov 4 '10 at 16:49
• @Derek: Please stop editing! I think I fixed it :-) (See revision 4 or 6) – Aryabhata Nov 11 '10 at 22:16