# Another Ramanujan's formula dealing with $\coth^{2}(5\pi)$

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.

Latest Update: I asked this question at MathOverflow (https://mathoverflow.net/q/173356/15540) and got a very beautiful answer.

• I suspect the Abel-Plana formula might be applicable to this series. – David H Jun 28 '14 at 19:29
• @DavidH: can you indicate how, as I am unable to see myself. – Paramanand Singh Jun 30 '14 at 3:57

This is just an idea. I write the series as:$$\sum_{n = 1}^{\infty}\frac{n^{5}q^{2n}}{1 - q^{2n}}\cdot\frac{1}{2500 + n^{4}}= \sum_{n = 1}^{\infty}\frac{n^{4}q^{2n}}{1 - q^{2n}}\cdot\frac{n}{2500 + n^{4}}$$ As you noticed: $$n^{4} + 2500 = (n^{2} + 50)^{2} - 100n^{2} = (n^{2} - 10n + 50)(n^{2} + 10n + 50)$$ Then \begin{aligned}S &= \sum_{n = 1}^{\infty}\left( \frac{1}{n^{2} - 10n + 50}-\frac{1}{n^{2} + 10n + 50} \right)\\ &= \sum_{n = 1}^{\infty} \left( \frac{1}{n^{2} - 10n + 50}-\frac{1}{(n+10)^{2} - 10(n+10) + 50} \right)\\ &= \sum_{n = 1}^{\infty} \left( \frac{1}{n^{2} - 10n + 50} \right)-\sum_{n = 10}^{\infty} \left( \frac{1}{(n-10)^{2} - 10(n-10) + 50} \right)\\ &=\sum_{n = 1}^{10} \left( \frac{1}{n^{2} - 10n + 50} \right)\\ &=\frac{4118807}{13138450}\end{aligned} And hence $$\sum_{n = 1}^{\infty}\frac{n}{2500 + n^{4}}=\frac{4118807}{262769000}$$ The fraction $$\frac{4118807}{262769000}=\frac{3\cdot4118807}{500\cdot1576614}=\frac{12356421}{500\cdot1576614}$$ Thus $$\frac{123826979}{6306456}=\frac{123826979}{4\cdot1576614}=\frac{15478372375}{12356421}\cdot\sum_{n = 1}^{\infty}\frac{n}{2500 + n^{4}}$$ We can write: $$\sum_{n = 1}^{\infty}\frac{n}{2500 + n^{4}} \left( \frac{15478372375}{12356421}-\frac{n^{4}}{e^{2 \pi n}-1}\right)=\frac{25}{4} \pi \coth^{2}(5 \pi)$$
• If you check revisions, you will find that you wrote 4118808. I just multiplied by $3$ and wrote $24$ instead of $21$. I did not check that part. – Paramanand Singh Jun 29 '14 at 17:25
• I have checked and the sum comes to be $4118807/13138450$ and I have updated post to reflect it. – Paramanand Singh Jun 30 '14 at 3:56