A coth series

Prove that

$\sum_{n=1}^{\infty} \frac{\coth n \pi}{n^7} = \frac{19 \pi^7}{56700}$

Solution

We recall the Fourier expansion of $\coth$ .

(1) $\begin{equation*}\pi \coth \pi a = \frac{1}{a} + \sum_{m=1}^{\infty} \frac{2a}{m^2+a^2} \end{equation*}$

Thus

$\begin{align*} \sum_{n=1}^{\infty} \frac{\coth n \pi}{n^7} &= \sum_{n=1}^{\infty} \frac{1}{n^7} \left [ \frac{1}{n \pi} + \frac{1}{\pi} \sum_{m=1}^{\infty} \frac{2n}{m^2+n^2} \right ] \\ &= \frac{1}{\pi} \zeta(8) + \frac{2}{\pi} \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^6 \left ( m^2+n^2 \right )}\\ &= \frac{\pi^7}{9450} + \frac{2}{\pi}\frac{13 \pi^8}{113400} \\ &= \frac{19 \pi^7}{56700} \end{align*}$

because

$\begin{aligned} \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{k^6 \left ( n^2 +k^2 \right )} &= \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{n^2+k^2-k^2}{k^6n^2 \left ( n^2+k^2 \right ) } \\ &= \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \left ( \frac{1}{k^6 n^2} - \frac{1}{k^4 n^2 \left ( n^2+k^2 \right )} \right )\\ &= \zeta(6) \zeta(2) - \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{k^4 n^2 \left ( n^2+k^2 \right ) }\\ &=\zeta(6) \zeta(2) - \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{n^2+k^2-k^2}{k^4 n^4 \left ( n^2+k^2 \right )} \\ &=\zeta(6) \zeta(2) - \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{k^4 n^4} \\ &\quad \quad \quad + \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{k^2 n^4 \left ( n^2+k^2 \right )}\\ &= \zeta(6) \zeta(2) - \zeta(4)^2 + \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{k^2 n^6} - \\ &\quad \quad \quad - \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^6 \left ( n^2+k^2 \right )} \\ &= \frac{1}{2} \left ( \zeta(2) \zeta(6) - \zeta^2(4) + \zeta(2)\zeta(6) \right ) \\ &= \frac{13 \pi^8}{113400} \end{aligned}$

A coth series

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