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Question Number 138163 by mnjuly1970 last updated on 10/Apr/21

$$........nice...........calculus.....$$$$provethat::$$$$\mathrm{\Psi }=\underset{n=1}{\sum ^{\mathrm{\infty }}}\left(\frac{1}{n^2{\pi }^2+1}\right)\stackrel{???}{=}\frac{1}{e^2-1}$$$$.............$$

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

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2+{\varphi }^2}=\frac{\pi }{2\mathrm{\varnothing }}coth\left(\pi \mathrm{\varnothing }\right)-\frac{1}{2{\mathrm{\varnothing }}^2}$$$$\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2{\pi }^2+1}=\frac{1}{{\pi }^2}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2+{\left(\frac{1}{\pi }\right)}^2}=\frac{1}{{\pi }^2}(\frac{{\pi }^2}{2}coth\left(1\right)-\frac{{\pi }^2}{2})$$$$=\frac{1}{2}(\frac{e^2+1}{e^2-1}-1)=\frac{1}{e^2-1}=\frac{\frac{1}{e}}{e-\frac{1}{e}}$$

Commented by mnjuly1970 last updated on 10/Apr/21

$$greateful..mr..payan...$$

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