TITLE: Infinite series $n^7/(\exp(2\pi n)-1)$
QUESTION [18 upvotes]: I found an interesting topic on this site with regards to the series I am trying to evaluate:
Summing $\frac{1}{e^{2\pi}-1} + \frac{2}{e^{4\pi}-1} + \frac{3}{e^{6\pi}-1} + \cdots \text{ad inf}$
I was wondering if there is a closed form for even m when we have:
$$\sum_{n=1}^{\infty}\frac{n^{2m-1}}{e^{2\pi n}-1}$$
I have the series $$\sum_{n=1}^{\infty}\frac{n^{7}}{e^{2\pi n}-1},$$
I am trying to evaluate. 
The aforementioned thread mentions that if m > 1 and odd, then we can use
$$\frac{B_{2m}}{4m}$$ to find the sum. 
But, if m is even, the formula omits and error term.
Does anyone have info on this error term or how to evaluate my series or others that involve even m?. 
I noticed that when m=1, there is an error term of $$-\frac{1}{8\pi}$$
The error appears to get smaller the larger m, and thus the power of n, becomes.
Thanks to all.

REPLY [3 votes]: The sums you want all appear in $\displaystyle E_{4k}(q)=1-\frac{4k}{B_{2k}}\sum_{n \ge 1}\frac{n^{4k-1} q^n}{1-q^n}\,$, with $q=e^{-2\pi}$ and $B_{2k}$ the ${2k}\,$th Bernoulli number. By the recurrence relation from Wikipedia (which uses $d_k=(4k+6)E_{2k+4}k!\zeta(2k+4)$), all $E_{4k\ge 8}$ can be expressed in terms of $E_4$ and $E_6$ (but $E_6(e^{-2\pi})\,$=$0$, so you only need to consider $E_4$).
Now to find $E_4$, plug in $\tau=i\;$ to the equation for the modular discriminant$$g_2^3(\tau)-27g_3^2(\tau)=(2\pi)^{12}e^{2i\pi\tau}\prod_{k\ge 1}(1-e^{2ik\pi\tau})^{24}$$and use the well known value $\prod_{k\ge 1}(1-e^{-2k\pi})=\frac{e^{\pi/12}\Gamma(1/4)}{2\pi^{3/4}}$ along with $g_2(i)=120\zeta(4)E_4(e^{-2\pi})$ and $g_3(i)=0$ (N.B. $g_3(i)=280\zeta(6)E_6(e^{-2\pi})$)
Finally, the recurrence relations simplify to $E_{4n}=A_n E_4^n$, where $A_n$ is a rather unwieldy rational number (the first few though can be figured easily from the Wikipedia link above, under the subsection "Products of Eisenstein series").
(Edit) I've confirmed the first two sums are $\sum_{k\ge1} \frac{k^3}{e^{2\pi k}-1}=-\frac1{240}+\color{maroon}{\frac{\Gamma^8(1/4)}{5120\pi^6}}$and$\sum_{k\ge1} \frac{k^7}{e^{2\pi k}-1}=-\frac1{480}+\color{maroon}{\frac{3\Gamma^{16}(1/4)}{655360\pi^{12}}}$, but I didn't get consistent results with Maple for the next sum.
(There shouldn't be any weird spaces now.)