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Question Number 42266 by maxmathsup by imad last updated on 21/Aug/18

$$findthevalueof\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n^2+1}and\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^2+1}.$$

Commented by maxmathsup by imad last updated on 22/Aug/18

$$wehaveprovedthat$$$$e^{-\mid x\mid }=\frac{1-e^{-\pi }}{\pi }+\frac{2}{\pi }\sum _{n=1}^{\mathrm{\infty }}\frac{1-{(-1)}^n{e}^{-\pi }}{n^2+1}cos\left(nx\right)$$$$x=0\Rightarrow 1=\frac{1-e^{-\pi }}{\pi }+\frac{2}{\pi }\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2+1}-\frac{2e^{-\pi }}{\pi }\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^2+1}let$$$$s=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2+1}andw=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^2+1}\Rightarrow$$$$1-\frac{1-e^{-\pi }}{\pi }=\frac{2}{\pi }s-\frac{2}{\pi }{e}^{-\pi }w\Rightarrow \frac{\pi }{2}\left(\frac{\pi -1+e^{-\pi }}{\pi }\right)=s-e^{-\pi }w\Rightarrow$$$$s-e^{-\pi }w=\frac{\pi -1+e^{-\pi }}{2}$$$$x=\pi \Rightarrow e^{-\pi }=\frac{1-e^{-\pi }}{\pi }+\frac{2}{\pi }\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^2+1}-\frac{2e^{-\pi }}{\pi }\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2+1}\Rightarrow$$$$e^{-\pi }-\frac{1-e^{-\pi }}{\pi }=\frac{2}{\pi }w-\frac{2e^{-\pi }}{\pi }s\Rightarrow \frac{\pi }{2}\left(\frac{\pi e^{-\pi }-1+e^{-\pi }}{\pi }\right)=-2e^{-\pi }s+w\Rightarrow$$$$\frac{\pi e^{-\pi }-1+e^{-\pi }}{2}=-2e^{-\pi }s+wwegetthesystem$$$$\mathbf{s}-{\mathbf{e}}^{-\pi }w=\frac{\pi -1+e^{-\pi }}{2}and-2e^{-\pi }s+w=\frac{(\pi +1)e^{-\pi }}{2}\Rightarrow$$$$-2e^{-2\pi }s+e^{-\pi }w=\frac{(\pi +1)e^{-2\pi }}{2}\Rightarrow$$$$(1-2e^{-2\pi })s=\frac{\pi -1+e^{-\pi }+(\pi +1)e^{-2\pi }}{2}\Rightarrow$$$$s=\frac{\pi -1+e^{-\pi }+(\pi +1)e^{-2\pi }}{2(1-2e^{-2\pi })}andw=\frac{(\pi +1)e^{-\pi }}{2}+2e^{-\pi }s.$$

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