Ramanujan mentions in one of his letters to Hardy that $$\frac{1^{5}}{e^{2\pi} - 1}\cdot\frac{1}{2500 + 1^{4}} + \frac{2^{5}}{e^{4\pi} - 1}\cdot\frac{1}{2500 + 2^{4}} + \cdots = \frac{123826979}{6306456} - \frac{25\pi}{4}\coth^{2}(5\pi)$$ If we put $q = e^{-\pi}$ we can see that the series is given by $$\sum_{n = 1}^{\infty}\frac{n^{5}q^{2n}}{1 - q^{2n}}\cdot\frac{1}{2500 + n^{4}}$$ While I am aware of the sum $$\sum_{n = 1}^{\infty}\frac{n^{5}q^{2n}}{1 - q^{2n}} = \frac{1 - R(q^{2})}{504}$$ and the Ramanujan function $R(q^{2})$ can be expressed in terms of $k, K$ as $$R(q^{2}) = \left(\frac{2K}{\pi}\right)^{6}(1 + k^{2})(1 - 2k^{2})\left(1 - \frac{k^{2}}{2}\right)$$ (see the derivation of this formula here). For $q = e^{-\pi}$ we have $k = 1/\sqrt{2}$ so that $R(q^{2}) = 0$ and hence $\sum_{n = 1}^{\infty}n^{5}q^{2n}/(1 - q^{2n}) = 1/504$. But getting the factor $1/(2500 + n^{4})$ seems really difficult.

Any ideas on whether we can get this factor by integration/differentiation (plus some algebraic games) from the series $\sum n^{5}q^{2n}/(1 - q^{2n})$?

**Further Update**: We have $$n^{4} + 2500 = (n^{2} + 50)^{2} - 100n^{2} = (n^{2} - 10n + 50)(n^{2} + 10n + 50)$$ so that $$n^{4} + 2500 = (n - 5 - 5i)(n - 5 + 5i)(n + 5 - 5i)(n + 5 + 5i)$$ so I believe we can do a partial fraction decomposition of $1/(n^{4} + 2500)$ but still I need to find a way to sum $\sum n^{5}q^{2n}/(1 - q^{2n})\cdot 1/(n + a)$ i.e. the problem is now simplified to getting a linear factor like $1/(n + a)$ somehow.