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Question Number 131580 by Dwaipayan Shikari last updated on 06/Feb/21

$$\mathbf{Prove}\mathbf{or}\mathbf{disprove}$$$$\underset{\mathit{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{({\mathit{n}}^2+97)}^2}=\frac{{\mathit{\pi }}^2}{97{({\mathit{e}}^{\mathit{\pi }\sqrt{97}}-e^{-\mathit{\pi }\sqrt{97}})}^2}+\frac{\mathit{\pi }}{388}.\frac{e^{2\mathit{\pi }\sqrt{97}}+1}{{\mathit{e}}^{2\mathit{\pi }\sqrt{97}}-1}+\frac{37635}{37636}-\frac{1}{388\sqrt{97}}$$

Commented by Dwaipayan Shikari last updated on 06/Feb/21

$$Ihavethisresultbutnotsure..$$

Commented by mindispower last updated on 09/Feb/21

$$seemexacte$$$$deducethis$$$$starting$$$$withe$$$$cot\left(x\right)=\frac{1}{x}+2\sum _{n⩾1}\frac{x}{x^2-{\pi }^2{n}^2}$$$$coth\left(x\right)=-\frac{1}{x}-2\mathrm{\Sigma }\frac{x}{x^2+n^2{\pi }^2}$$$$\frac{1}{y}coth\left(\pi y\right)=-\frac{1}{\pi y^2}-\frac{2}{\pi }\mathrm{\Sigma }\frac{1}{y^2+n^2}$$$$\mathrm{\Sigma }\frac{1}{(y^2+n^2)}=\frac{\pi }{2}(\frac{1}{\pi y^2}-\frac{coth\left(\pi y\right)}{y})$$$$\Rightarrow \sum _{n⩾1}\frac{2y}{{(y^2+n^2)}^2}=-\frac{1}{y^3}+\frac{\pi }{2y^2}coth\left(\pi y\right)-\frac{{\pi }^2}{2y}(-1+{coth}^2\left(\pi y\right))$$

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