Otro Ramanujan la fórmula de tratar con $\coth^{2}(5\pi)$

Question

Ramanujan menciona en una de sus cartas a Hardy que $$\frac{1^{5}}{e^{2\pi} - 1}\cdot\frac{1}{2500 + 1^{4}} + \frac{2^{5}}{e^{4\pi} - 1}\cdot\frac{1}{2500 + 2^{4}} + \cdots = \frac{123826979}{6306456} - \frac{25\pi}{4}\coth^{2}(5\pi)$$ 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 $\sum_{n = 1}^{\infty}n^{5}p^{2n}/(1 - p^{2n}) = 1/504$. But getting the factor $1/(2500 + n^{4})$ parece realmente difícil.

Alguna idea sobre si podemos obtener este factor de integración/diferenciación (además de algunos algebraicas juegos) de la serie $\sum n^{5}q^{2n}/(1 - q^{2n})$?

Actualización: Hemos $$n^{4} + 2500 = (n^{2} + 50)^{2} - 100n^{2} = (n^{2} - 10n + 50)(n^{2} + 10n + 50)$$ 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 $\suma de n^{5}p^{2n}/(1 - p^{2n})\cdot 1/(n + a)$ i.e. the problem is now simplified to getting a linear factor like $1/(n + a)$ de alguna manera.

Última Actualización: hice esta pregunta en MathOverflow (http://mathoverflow.net/q/173356/15540) y tiene una muy hermosa respuesta.

Answer 1

Esto es sólo una idea. Escribo la serie como:$$\sum_{n = 1}^{\infty}\frac{n^{5}p^{2n}}{1 - p^{2n}}\cdot\frac{1}{2500 + n^{4}}= \sum_{n = 1}^{\infty}\frac{n^{4}p^{2n}}{1 - p^{2n}}\cdot\frac{n}{2500 + n^{4}} $$ As you noticed: $$n^{4} + 2500 = (n^{2} + 50)^{2} - 100n^{2} = (n^{2} - 10n + 50)(n^{2} + 10n + 50)$$ Then $$\begin{aligned}S &= \sum_{n = 1}^{\infty}\left( \frac{1}{n^{2} - 10n + 50}-\frac{1}{n^{2} + 10n + 50} \right)\\ &= \sum_{n = 1}^{\infty} \left( \frac{1}{n^{2} - 10n + 50}-\frac{1}{(n+10)^{2} - 10(n+10) + 50} \right)\\ &= \sum_{n = 1}^{\infty} \left( \frac{1}{n^{2} - 10n + 50} \right)-\sum_{n = 10}^{\infty} \left( \frac{1}{(n-10)^{2} - 10(n-10) + 50} \right)\\ &=\sum_{n = 1}^{10} \left( \frac{1}{n^{2} - 10n + 50} \right)\\ &=\frac{4118807}{13138450}\end{alineados}$$ 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: $$ \sum_{n = 1}^{\infty}\frac{n}{2500 + n^{4}} \left( \frac{15478372375}{12356421}-\frac{n^{4}}{e^{2 \pi n}-1}\right)=\frac{25}{4} \pi \coth^{2}(5 \pi)$$