Closed form for $\sum\limits_{n=1}^{\infty}\frac{n^{4k-1}}{e^{n\pi}-1}-16^k\sum\limits_{n=1}^{\infty}\frac{n^{4k-1}}{e^{4n\pi}-1}$ We took the idea from this Ramanujan's identity
$$\frac{1^{13}}{e^{2\pi}-1}+\frac{2^{13}}{e^{4\pi}-1}+\frac{3^{13}}{e^{6\pi}-1}+\cdots=\frac{1}{24}$$
A few examples of Ramanujan-type identities
$$\sum_{n=1}^{\infty}\frac{n^3}{e^{n\pi}-1}-16\sum_{n=1}^{\infty}\frac{n^3}{e^{4n\pi}-1}=\frac{1}{16}$$
$$\sum_{n=1}^{\infty}\frac{n^7}{e^{n\pi}-1}-16^2\sum_{n=1}^{\infty}\frac{n^7}{e^{4n\pi}-1}=\frac{17}{32}$$
$$\sum_{n=1}^{\infty}\frac{n^{11}}{e^{n\pi}-1}-16^3\sum_{n=1}^{\infty}\frac{n^{11}}{e^{4n\pi}-1}=\frac{691}{16}$$
Thus, for every integer $k\ge1$, we define

$$F(k)=\sum_{n=1}^{\infty}\frac{n^{4k-1}}{e^{n\pi}-1}-16^k\sum_{n=1}^{\infty}\frac{n^{4k-1}}{e^{4n\pi}-1}$$

Question: What is a closed form for this Ramanujan type function $F(k)$?
We believe that this involves only some rational closed form.
 A: Just for fun here is a way of obtaining a closed form expression for the sum (which turns out to be the correct one) by manipulating formulas with a complete disregard for divergent sums (the more divergent the better). 

We start with the generating function for the Bernoulli numbers
$$\frac{x^{4k-1}}{e^{x}-1} = \sum_{j\geq 0} \frac{B_j}{j!}x^{j+4k-2}$$
Taking $x=n$ and $x=4n$ respectively we obtain
$$\frac{n^{4k-1}}{e^{n\pi}-1}-16^{k}\frac{n^{4k-1}}{e^{4n\pi}-1}  = \frac{1}{\pi}\sum_{j\geq 0} \frac{B_j}{j!}\left[1-4^{j+2k-1}\right]\pi^j \cdot \frac{1}{n^{2-j-4k}}$$
Now by summing this equation over $n$, using the definition of $\zeta$-function, we get
$$\frac{1}{\pi}\sum_{j\geq 0} \frac{B_j}{j!}\left[1-4^{j+2k-1}\right]\pi^j\zeta(2-j-4k)$$
Note that the $\zeta$-argument is negative so the sum is as divergent as it's possible to get (each term is a divergent sum), but by interpretting it using the analytical continuation of $\zeta(s)$ we can try to extract some meaning from it. The analytical continuation satisfy $\zeta(2-j-4k) = 0$ when $j$ is even and since $B_{j} = 0$ when $j$ is odd (apart from the term $j=1$ having $B_1 = - \frac{1}{2}$) the formula above reduces to
$$-\frac{1}{2}\left[1-4^{2k}\right]\zeta(1-4k)$$
Finally, by using the $\zeta$ functional equation and the relation for $\zeta(2n)$ when $n$ is an integer this simplifies to
$$\frac{B_{4k}}{8k}\left(1-4^{2k}\right)$$
Remarkably this calculation gives us the correct result (as it often does).
