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Question Number 123067 by Khalmohmmad last updated on 22/Nov/20

$$\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{2}{n^2-1}=?$$

Answered by Dwaipayan Shikari last updated on 22/Nov/20

$$2\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2-1}=\sum ^{\mathrm{\infty }}\frac{1}{n-1}-\sum ^{\mathrm{\infty }}\frac{1}{n+1}=(1+\frac{1}{2}+\frac{1}{3}+..)-(\frac{1}{3}+\frac{1}{4}+..)=1+\frac{1}{2}$$$$=\frac{3}{2}$$

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