RAMANUJAN´S INFINITE SUM

Today we will proof the following two beautiful results found here. The first one is known to be an infinite sum proved by Ramanujan to Hardy, the second I am not sure.


\sum_{n=1}^{\infty} \frac{\operatorname{coth} \pi n x+x^{2} \operatorname{coth} \frac{\pi n}{x}}{n^{3}}=\frac{\pi^{3}}{90 x}\left(1+5 x^{2}+x^{4}\right)(1)

\sum_{n=1}^{\infty} \frac{\operatorname{coth} \pi n x-x^{4} \operatorname{coth} \frac{\pi n}{x}}{n^{5}}=\frac{\pi^{5}}{1890 x}\left(1-x^2 \right)\left(2x^4+9 x^{2}+2\right)(2)

First, recall the partial fraction expansion of the hyperbolic cotangent

\coth\left(\pi x \right)&=\frac{1}{x \pi}+\frac{1}{\pi}\sum_{k=1}^{\infty}\frac{2x}{k^2+x^2}

Then we first sum becomes

\begin{aligned} &S= \sum_{n=1}^{\infty} \frac{\operatorname{coth} \pi n x+x^{2} \operatorname{coth} \frac{\pi n}{x}}{n^{3}}\\ &=\sum_{n=1}^{\infty}\frac{1}{n^3}\left(\frac{1}{\pi nx }+\frac{2}{\pi}\sum_{k=1}^{\infty}\frac{ nx}{k^2+( nx)^2}\right)+x^2\sum_{n=1}^{\infty}\frac{1}{n^3}\left(\frac{x}{\pi n }+\frac{2}{\pi }\sum_{k=1}^{\infty}\frac{ \frac{n}{x}}{k^2+\left( \frac{n}{x}\right)^2}\right)\\ &=\frac{1}{\pi }\left(\frac{1}{x}+x^3 \right)\sum_{n=1}^{\infty}\frac{1}{n^4}+\frac{2}{\pi}\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{n^2}\left( \frac{x}{k^2+( n x )^2}+\frac{ x}{k^2+\left( \frac{n}{x}\right)^2}\right)\\ &=\frac{1}{\pi }\left(\frac{1}{x}+x^3 \right)\zeta(4)+\frac{2}{\pi}\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{n^2}\left( \frac{x}{k^2+( n x )^2}+\frac{ x}{k^2+\left( \frac{n}{x}\right)^2}\cdot\frac{x^2}{x^2}\right)\\ &=\frac{\pi^3}{90 }\left(\frac{1}{x}+x^3 \right)+\frac{2}{\pi}\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{n^2}\left( \frac{x}{k^2+( n x )^2}+\frac{ x^3}{n^2+\left(kx\right)^2}\right)\\ &=\frac{\pi^3}{90 }\left(\frac{1}{x}+x^3 \right)+\frac{2x}{\pi}\left(\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{n^2} \frac{1}{k^2+( n x )^2}+\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{n^2}\frac{ x^2}{n^2+\left(kx\right)^2}\right)\\ \end{aligned}

Since n and k are dummy variables, we can re-index the second double sum exchanging n by k and vice versa, then

\begin{aligned} &=\frac{\pi^3}{90 }\left(\frac{1}{x}+x^3 \right)+\frac{2x}{\pi}\left(\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{n^2} \frac{1}{k^2+( n x )^2}+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{k^2}\frac{ x^2}{k^2+\left(nx\right)^2}\right)\\ &=\frac{\pi^3}{90 }\left(\frac{1}{x}+x^3 \right)+\frac{2x}{\pi}\left(\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{k^2}{k^2}\cdot\frac{1}{n^2} \frac{1}{k^2+( n x )^2}+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{n^2}{n^2}\cdot\frac{1}{k^2}\frac{ x^2}{k^2+\left(nx\right)^2}\right)\\ &=\frac{\pi^3}{90 }\left(\frac{1}{x}+x^3 \right)+\frac{2x}{\pi}\left(\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{k^2}\cdot\frac{1}{n^2} \frac{k^2}{k^2+( n x )^2}+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{n^2}\cdot\frac{1}{k^2}\frac{ (nx)^2}{k^2+\left(nx\right)^2}\right)\\ &=\frac{\pi^3}{90 }\left(\frac{1}{x}+x^3 \right)+\frac{2x}{\pi}\left(\sum_{n=1}^{\infty}\frac{1}{n^2}\sum_{k=1}^{\infty}\frac{1}{k^2}\cdot \frac{k^2+(nx)^2}{k^2+( n x )^2}\right)\\ &=\frac{\pi^3}{90 }\left(\frac{1}{x}+x^3 \right)+\frac{2x}{\pi}\left(\zeta(2)\right)^2\\ &=\frac{\pi^3}{90 }\left(\frac{1}{x}+x^3 \right)+\frac{2x}{\pi}\frac{\pi^4}{36}\\ &=\frac{\pi^3}{90 }\left(\frac{1}{x}+x^3 +5x\right) \,\blacksquare\\ \end{aligned}

For the second sum

\begin{aligned} &S= \sum_{n=1}^{\infty} \frac{\operatorname{coth} \pi n x-x^{4} \operatorname{coth} \frac{\pi n}{x}}{n^{5}}\\ &=\sum_{n=1}^{\infty}\frac{1}{n^5}\left(\frac{1}{\pi nx }+\frac{2}{\pi}\sum_{k=1}^{\infty}\frac{ nx}{k^2+( nx)^2}\right)-x^4\sum_{n=1}^{\infty}\frac{1}{n^5}\left(\frac{x}{\pi n }+\frac{2}{\pi }\sum_{k=1}^{\infty}\frac{ \frac{n}{x}}{k^2+\left( \frac{n}{x}\right)^2}\right)\\ &=\frac{1}{\pi }\left(\frac{1}{x}-x^5 \right)\sum_{n=1}^{\infty}\frac{1}{n^6}+\frac{2}{\pi}\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{n^4}\left( \frac{x}{k^2+( n x )^2}-\frac{ x^3}{k^2+\left( \frac{n}{x}\right)^2}\right)\\ &=\frac{1}{\pi }\left(\frac{1}{x}-x^5 \right)\zeta(6)+\frac{2x}{\pi}\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{n^4}\left( \frac{1}{k^2+( n x )^2}-\frac{ x^2}{k^2+\left( \frac{n}{x}\right)^2}\cdot\frac{x^2}{x^2}\right)\\ &=\frac{1}{\pi }\left(\frac{1}{x}-x^5 \right)\zeta(6)+\frac{2x}{\pi}\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{n^4}\left( \frac{1}{k^2+( n x )^2}-\frac{ x^4}{n^2+\left(kx\right)^2}\right)\\ &=\frac{1}{\pi }\left(\frac{1}{x}-x^5 \right)\zeta(6)+\frac{2x}{\pi}\left(\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{n^4} \frac{1}{k^2+( n x )^2}-\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{n^4}\frac{ x^4}{n^2+\left(kx\right)^2}\right)\\ \end{aligned}

\begin{aligned} &=\frac{1}{\pi }\left(\frac{1}{x}-x^5 \right)\zeta(6)+\frac{2x}{\pi}\left(\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{n^4} \frac{1}{k^2+( n x )^2}-\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{k^4}\frac{ x^4}{k^2+\left(nx\right)^2}\right)\\ &=\frac{1}{\pi }\left(\frac{1}{x}-x^5 \right)\zeta(6)+\frac{2x}{\pi}\left(\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{k^4}{k^4}\cdot\frac{1}{n^4} \frac{1}{k^2+( n x )^2}-\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{n^4}{n^4}\cdot\frac{1}{k^4}\frac{ x^4}{k^2+\left(nx\right)^2}\right)\\ &=\frac{1}{\pi }\left(\frac{1}{x}-x^5 \right)\zeta(6)+\frac{2x}{\pi}\left(\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{k^4}\cdot\frac{1}{n^4} \frac{k^4}{k^2+( n x )^2}-\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{n^4}\cdot\frac{1}{k^4}\frac{ (nx)^4}{k^2+\left(nx\right)^2}\right)\\ &=\frac{1}{\pi }\left(\frac{1}{x}-x^5 \right)\zeta(6)+\frac{2x}{\pi}\left(\sum_{n=1}^{\infty}\frac{1}{n^4}\sum_{k=1}^{\infty}\frac{1}{k^4}\cdot \frac{k^4-(nx)^4}{k^2+( n x )^2}\right)\\ &=\frac{1}{\pi }\left(\frac{1}{x}-x^5 \right)\zeta(6)+\frac{2x}{\pi}\left(\sum_{n=1}^{\infty}\frac{1}{n^4}\sum_{k=1}^{\infty}\frac{1}{k^4}\cdot \frac{(k^2+(nx)^2)(k^2-(nx)^2)}{k^2+( n x )^2}\right)\\ &=\frac{1}{\pi }\left(\frac{1}{x}-x^5 \right)\zeta(6)+\frac{2x}{\pi}\left(\sum_{n=1}^{\infty}\frac{1}{n^4}\sum_{k=1}^{\infty}\frac{1}{k^4}\cdot (k^2-(nx)^2)\right)\\ &=\frac{1}{\pi }\left(\frac{1}{x}-x^5 \right)\zeta(6)+\frac{2x}{\pi}\left(\sum_{n=1}^{\infty}\frac{1}{n^4}\sum_{k=1}^{\infty}\frac{k^2}{k^4}-x^2\sum_{n=1}^{\infty}\frac{n^2}{n^4}\sum_{k=1}^{\infty}\frac{1}{k^4}\right)\\ &=\frac{1}{\pi }\left(\frac{1}{x}-x^5 \right)\zeta(6)+\frac{2x}{\pi}\left(1-x^2 \right)\left(\zeta(2) \zeta(4)\right)\\ &=\left(\frac{1}{x}-x^5 \right)\frac{\pi^5}{945} +x\left(1-x^2 \right)\left(\frac{\pi^5}{270}\right)\\ &=\frac{1}{x}\left(1-x^6 \right)\frac{\pi^5}{945} +x\left(1-x^2 \right)\left(\frac{\pi^5}{270}\right)\\ &=\frac{\pi^{5}}{1890 x}\left[2\left(1-x^6\right)+7x^2\left(1-x^2 \right)\right]\\ &=\frac{\pi^{5}}{1890 x}\left(1-x^2 \right)\left(2x^4+9 x^{2}+2\right)\,\blacksquare \end{aligned}

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