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.


Using (see entry 24 on page 291 in Ramanujan's Notebooks II: http://www.plou) $$\frac{\pi e^{-2\pi z}}{2z[\cosh{(2\pi z)}-\cos{(2\pi z)}]}= \frac{1}{8\pi z^3}-\frac{1}{4z^2}+\frac{\pi}{4z}-\sum\limits_{n=1}^\infty \frac{1}{z^2+(z+n)^2}$$ $$+4z\sum\limits_{n=1}^\infty \frac{n}{(e^{2\pi n}-1) (4z^4+n^4)},$$ and $$\frac{n}{4z^4+n^4}=\frac{1}{4z^4}\left (n-\frac{n^5}{4z^4+n^4}\right ),$$ we get $$\sum\limits_{n=1}^\infty \frac{n^5}{(e^{2\pi n}-1) (4z^4+n^4)}=\frac{1}{8\pi}-\frac{z}{4}+\frac{\pi z^2}{4}- \sum\limits_{n=1}^\infty\frac{z^3}{z^2+(z+n)^2}$$ $$+\sum\limits_{n=1}^\infty \frac{n}{e^{2\pi n}-1}-\frac{\pi z^2e^{-2\pi z}} {2[\cosh{(2\pi z)}-\cos{(2\pi z)}]}.$$ Let us substitute $z=5i$ in this equation. Then $$\frac{\pi z^2e^{-2\pi z}}{2[\cosh{(2\pi z)}-\cos{(2\pi z)}]}= \frac{-25\pi}{2(1-\cosh{(10\pi)})}=\frac{25\pi}{4\sinh^2{(5\pi)}}=$$ $$ \frac{25\pi}{4}\coth^2{(5\pi)}-\frac{25\pi}{4}.$$ Therefore we get $$\sum\limits_{n=1}^\infty \frac{n^5}{(e^{2\pi n}-1)(2500+n^4)}= \frac{1}{8\pi}-\frac{5i}{4}+125i\sum\limits_{n=1}^\infty \frac{1}{(5i+n)^2-25}$$ $$+\sum\limits_{n=1}^\infty\frac{n}{e^{2\pi n}-1} -\frac{25\pi}{4}\coth^2{(5\pi)}. \tag{1}$$ Now (see page 6 in David M. Bradley, Ramanujan's formula for the logarithmic
derivative of the gamma function: http://arxiv.org/abs/math/0505125) $$\sum\limits_{n=1}^\infty\frac{n}{e^{2\pi n}-1}= \frac{1}{24}-\frac{1}{8\pi}, \tag{2}$$ and $$\sum\limits_{n=1}^\infty\frac{1}{(5i+n)^2-25}=\frac{1}{10} \sum\limits_{n=1}^\infty\left (\frac{1}{n-5+5i}-\frac{1}{n+5+5i}\right)=$$ $$ \frac{1}{10}\left (\sum\limits_{n=-9}^\infty\frac{1}{n+5+5i}- \sum\limits_{n=1}^\infty\frac{1}{n+5+5i}\right)=\frac{1}{10} \sum\limits_{n=-9}^0\frac{1}{n+5+5i}. \tag{3}$$ Therefore, calculating the finite sum in (3), $$125i\sum\limits_{n=1}^\infty\frac{1}{(5i+n)^2-25}= \frac{5i}{4}+\frac{20594035}{1051076}. \tag{4}$$ Substituting (2) and (4) in (1), we get the Ramanujan's formula, because $$\frac{20594035}{1051076}+\frac{1}{24}= \frac{123826979}{6306456}.$$

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    $\begingroup$ Thanks for very clear and detailed solution. +1. I had a copy of Ramanujan's Notebooks Vol 2 with me, but it seems I just could not think of this particular entry to connect with the current question. In fact your answer clearly shows that the connection is not so obvious and does require some real labor. $\endgroup$ – Paramanand Singh Jul 9 '14 at 16:01

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