14
$\begingroup$

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

  • 0
    This sum reappeared at this [MSE link](http://math.stackexchange.com/questions/392706/evaluating-sum-n-1-infty-fracne2-pi-n-1-using-the-inverse-melli).2014-01-29

1 Answers 1

25

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

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

  • 0
    @DerekJennings: How do you prove the generalization you have mentioned in the end of your answer? For some values of $m$ this can be established using the the Ramanujan's formulas for expressing these Eisenstein series in terms of $K, k$ and then putting $k = 1/\sqrt{2}$ corresponding to $q = e^{-\pi}$.2014-03-01