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Question Number 215516 by MrGaster last updated on 09/Jan/25

$$\mathrm{prove}:$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{\pi }^2{n}^2+1}=\frac{1}{e^2-1}$$

Answered by mr W last updated on 09/Jan/25

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{\pi }^2{n}^2+1}$$$$=\frac{1}{2i}\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{\pi n-i}-\frac{1}{\pi n+i})$$$$=\frac{1}{2\pi i}\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{n-\frac{i}{\pi }}-\frac{1}{\pi +\frac{i}{\pi }})$$$$=\frac{1}{2\pi i}[\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}-\gamma -\psi (1-\frac{i}{\pi })-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}+\gamma +\psi (1+\frac{i}{\pi })]$$$$=\frac{1}{2\pi i}[\psi (1+\frac{i}{\pi })-\psi (1-\frac{i}{\pi })]$$$$=\frac{1}{2\pi i}[\psi \left(\frac{i}{\pi }\right)+\frac{\pi }{i}-\psi \left(\frac{i}{\pi }\right)-\pi \cot i]$$$$=\frac{1}{2\pi i}(\frac{\pi }{i}-\pi \cot i)$$$$=\frac{1}{2\pi i}(\frac{\pi }{i}-\frac{\pi }{i}\coth 1)$$$$=\frac{1}{2}(\coth 1-1)$$$$=\frac{1}{2}(\frac{e+e^{-1}}{e-e^{-1}}-1)$$$$=\frac{e^{-1}}{e-e^{-1}}$$$$=\frac{1}{e^2-1}✓$$

Commented by ajfour last updated on 10/Jan/25

https://youtu.be/dvzKfeWOqyw?si=EggqbIQMc8QTN9F6 LCR Series circuit current and voltages discussed.

Commented by MrGaster last updated on 09/Jan/25

$$\mathrm{Than}\mathbb{k}\mathrm{you}\mathrm{sir}.$$

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