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How to calculate the closed form of the following Euler type Sums $$\sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{{n^p}\sinh \left( {n\pi } \right)}}} = ?$$ Here the harmonic number $H_n$ is defined by $${H_n} = \sum\limits_{j = 1}^n {\frac{1}{j}}$$.

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  • $\begingroup$ What makes you thinking that a closed form exists ? $\endgroup$ – Claude Leibovici Jun 10 '17 at 4:52
  • $\begingroup$ Since $\sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{{n^k}}}}$ have a closed form that $$\sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{{n^k}}}} = \frac{1}{2}\left\{ {\left( {k + 2} \right)\zeta \left( {k + 1} \right) - \sum\limits_{i = 1}^{k - 2} {\zeta \left( {k - i} \right)\zeta \left( {i + 1} \right)} } \right\}.$$ Hence, I think the sums above that it shoud be a similar closed form. $\endgroup$ – xuce1234 Jun 10 '17 at 4:56
  • $\begingroup$ Mi first temptation is to expand $\frac{1}{\sinh(\pi n)}$ as $$\frac{1}{\pi n}+\sum_{m\geq 1}(-1)^m\left(\frac{1}{n-mi}+\frac{1}{n+mi}\right)$$ then hope that $$\sum_{n\geq 1}\frac{H_n}{n^p(n+mi)}$$ is manageable in terms of Hurwitz zeta functions, as a natural extension of the $m=0$ case. $\endgroup$ – Jack D'Aurizio Jun 10 '17 at 17:36

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