Recently I have been playing around with series of the form
$$\sum_{k=1}^{\infty}\frac{k^{s}}{e^{kz}-1} = \sum_{k=1}^{\infty}\sigma_{s}(k)e^{-kz}$$
for $s \in \mathbb{Z}$ and where $\sigma_s(k)$ is the sum of divisors function of order $s$. These series have generated quite a bit of interest over the years, due in large part to some beautiful modular identities of Ramanujan. The most famous example being
$$\alpha^{-n}\left(\frac{1}{2}\zeta(2n+1)+\sum_{k=1}^{\infty}\frac{k^{2n-1}}{e^{2\alpha k}-1}\right) = \\ (-\beta)^n\left(\frac{1}{2}\zeta(2n+1)+\sum_{k=1}^{\infty}\frac{k^{2n-1}}{e^{2\beta k}-1}\right) – 2^{2n}\sum_{k=0}^{n+1}(-1)^k\frac{B_{2k}}{(2k)!}\frac{B_{2n+2-2k}}{(2n+2-2k)!}\alpha^{n+1-k}\beta^k$$
where $\alpha,\beta > 0, \alpha\beta=\pi^2$ and $B_k$ are the Bernoulli numbers and $\zeta(k)$ is the Riemann zeta function. As far as I know there aren't any similar relations or closed forms when $s \in 2\mathbb{Z}$. In my investigations I was able to find some approximation formula for general $s > 0$ but which unfortunately perform poorer and poorer as $s \rightarrow \infty$.
For instance, at $s=2$ we have
$$\sum_{k=1}^{\infty}\frac{k^2z}{e^{kz}-1} \approx \frac{2\zeta(3)}{z^2} – \frac{1}{2}-\frac{z}{24} -\sum_{j=0}^{N}B^{(2)}_{j+2}B_{j}\frac{z^{j}}{(j+2)!}$$
where $B^{(k)}_n$ are the Norlund polynomials.
I was excited to find this, but then unfortunately realized that since the sum on the right hand side diverges as $N \rightarrow \infty$ we can only achieve a finite number of accurate digits as the RHS approaches the left from below then surpasses it, growing without bound.
For instance letting $N=37$ we have
$$\sum_{k=1}^{\infty}\frac{k^2}{e^{k}-1} \approx 2 \zeta (3)-\frac{707928034947324016593079681811720894660110227517}{8567110474102926210628918330759216889856000000000}$$
with the right hand side being correct to the 14-th decimal place. This is about the best we can do with the above formula.
I am curious as to whether someone would be able to provide a better approximation. I am not very familiar with sort of thing… so maybe there is a standard way of achieving approximations like the one above?
Best Answer
I figured I would go ahead and provide an answer to my own question just in case someone else comes across this post and is interested in what else is known about formulae such as the one above.
Through my research I discovered two things. The first being that the above formula should be thought of as an asymptotic expansion. In particular, as $z \to 0$ we have
$$\sum_{k=1}^{\infty}\frac{k^2}{e^{kz}-1} = \frac{2\zeta(3)}{z^3} - \frac{41}{72z}-\frac{1}{24} -\sum_{j=1}^{N}B^{(2)}_{j+2}B_{j}\frac{z^{j-1}}{(j+2)!} + \mathcal{O}(\vert{z^{N}}\vert)$$
The second thing being, that there is some precedent for an identity of this sort. Namely, Wigert showed in 1916 that for any $N \geq 1$ the following asymptotic expansion holds as $z\to 0$ in any angle $\vert{\arg(z)}\vert<\pi/2$
$$\sum_{n=1}^{\infty}d(n)e^{-nz} = \frac{\gamma-\log(z)}{z}+\frac{1}{4}-\sum_{k=0}^{N-1}\frac{B^2_{2k+2}}{(2k+2)!(2k+2)}z^{2k+1}+\mathcal{O}(\vert{z^{2N}}\vert)$$
which when considering the identity
$$\sum_{n=1}^{\infty}\frac{x^n}{1-x^{n}} = \sum_{n=1}^{\infty}d(n)x^{n}$$
makes plain the connection between my identity and Wigert's.
My general conjecture is that for any $n,N \geq 1$ the following asymptotic expansion holds as $z\to 0$ in any angle $\vert{\arg(z)}\vert<\pi/2$
\begin{multline*} \sum_{k=1}^{\infty}\frac{k^{n+1}}{e^{kz}-1} = (-1)^{n}\sum_{k=0}^{n}k!S_2({n+1},{k+1})\sum_{j=0}^{k}B^{(k+1)}_j(-1)^{k-j}(k+1-j)\zeta(k+2-j)\frac{z^{j-k-2}}{j!} \\ + (-1)^{n+1}\sum_{k=0}^{n}k!S_2({n+1},{k+1})\sum_{j=0}^{N}B^{(k+1)}_{j+k+1}B_{j}\frac{z^{j-1}}{(j+k+1)!} +\mathcal{O}(\vert{z^{N}}\vert) \end{multline*}
where $S_2(n,k)$ are the Stirling numbers of the second kind.