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Question Number 143098 by mnjuly1970 last updated on 10/Jun/21

$$$$$$.....Prove....$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{\left(\frac{1}{\mathrm{s}inh\left(\pi n\right)}\right)}^2=\frac{1}{6}-\frac{1}{2\pi }...$$$$......$$

Answered by Dwaipayan Shikari last updated on 10/Jun/21

$$4\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{e^{2\pi n}}{{(e^{2\pi n}-1)}^2}$$$$\frac{\sin h\left(\pi x\right)}{\pi x}=\underset{n=1}{\prod ^{\mathrm{\infty }}}(1+\frac{x^2}{n^2})$$$$\Rightarrow coth\left(\pi x\right)-\frac{1}{\pi x}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{\frac{2x}{n^2}}{1+\frac{x^2}{n^2}}$$$$\Rightarrow \frac{e^{2\pi x}+1}{e^{2\pi x}-1}-\frac{1}{\pi x}=2\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x}{n^2+x^2}$$$$\Rightarrow 1+\frac{2}{e^{2\pi x}-1}-\frac{1}{\pi x}=2\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x}{(n^2+x^2)}$$$$\frac{-4\pi e^{2\pi x}}{{(e^{2\pi x}-1)}^2}+\frac{1}{\pi x^2}=-2\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n^2+x^2-2x}{{(n^2+x^2)}^2}$$$$\Rightarrow \underset{x=1}{\sum ^{\mathrm{\infty }}}\frac{e^{2\pi x}}{{(e^{2\pi x}-1)}^2}=\frac{1}{4}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{\pi }^2{x}^2}+\frac{1}{2}\underset{x=1}{\sum ^{\mathrm{\infty }}}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{(n^2+x^2)}-\frac{2x}{{(n^2+x^2)}^2}$$$$=\frac{1}{4}(.\frac{{\pi }^2}{{\pi }^{2}6})+\frac{1}{2}\mathrm{\Phi }=\frac{1}{24}+\mathrm{\Phi }...$$

Commented by Dwaipayan Shikari last updated on 10/Jun/21

$$Sorrysir.Ihavetried$$

Commented by mnjuly1970 last updated on 10/Jun/21

$$gratefulforyoureffortmrpayan..$$

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