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](http://paramanands.blogspot.com/2013/05/certain-lambert-series-identities-and-their-proof-via-trigonometry-part-2.html)). 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.



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



A sum by Ramanujan for $\coth^{2}(5\pi)$
Using (see entry 24 on page 291 in Ramanujan's Notebooks II: http://[www.plou](http://www.plouffe.fr/simon/math/Ramanujan%27s%20Notebooks%20II.pdf))
$$\frac{\pi e^{-2\pi z}}{2z[\cosh{(2\pi z)}-\cos{(2\pi z)}]}=
\frac{1}{8\pi z^3}-\frac{1}{4z^2}+\frac{\pi}{4z}-\sum\limits\_{n=1}^\infty
\frac{1}{z^2+(z+n)^2}$$ $$+4z\sum\limits\_{n=1}^\infty \frac{n}{(e^{2\pi n}-1)
(4z^4+n^4)},$$
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
$$\frac{20594035}{1051076}+\frac{1}{24}=
\frac{123826979}{6306456}.$$


