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Question Number 99894 by bachamohamed last updated on 23/Jun/20

$$$$$$\sum _{\mathit{n}⩾0}\frac{1}{{\mathit{n}}^2+1}=?$$

Answered by smridha last updated on 24/Jun/20

$$\mathit{s}\left(\mathit{a}\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{\mathit{n}}^2+{\mathit{a}}^2}=\frac{\mathit{\pi }\mathit{c}\mathit{o}\mathit{t}\mathit{h}\left(\mathit{\pi }\mathit{a}\right)}{2\mathit{a}}-\frac{1}{2{\mathit{a}}^2}$$$$\mathit{n}\mathit{o}\mathit{w}\mathit{s}\left(1\right)=\frac{\mathit{\pi }\mathit{c}\mathit{o}\mathit{t}\mathit{h}\left(\mathit{\pi }\right)}{2}-\frac{1}{2}\approx 1.0766...$$$$\mathit{n}\mathit{o}\mathit{w}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{\mathit{n}}^2+1}=1+\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2+1}\approx 2.0766....$$

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