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Question Number 114023 by Khalmohmmad last updated on 16/Sep/20

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

Commented by Dwaipayan Shikari last updated on 16/Sep/20

$$Diverges$$$$firstterm\frac{1}{n^2-1}\to \mathrm{\infty }(n=1)orundefined$$

Answered by mr W last updated on 16/Sep/20

$$\underset{\mathrm{n}=2}{\sum ^{\propto }}\frac{1}{{\mathrm{n}}^2-1}$$$$=\frac{1}{2}\underset{\mathrm{n}=2}{\sum ^{\propto }}(\frac{1}{n-1}-\frac{1}{n+1})$$$$=\frac{1}{2}(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...)$$$$=\frac{1}{2}\times \frac{3}{2}$$$$=\frac{3}{4}$$

Commented by Olaf last updated on 16/Sep/20

$$\mathrm{Erratum}!$$$${\mathrm{S}}_n=\frac{1}{2}(\underset{k=2}{\sum ^n}\frac{1}{k-1}-\underset{k=2}{\sum ^n}\frac{1}{k+1})$$$${\mathrm{S}}_n=\frac{1}{2}(\underset{k=1}{\sum ^{n-1}}\frac{1}{k}-\underset{k=3}{\sum ^{n+1}}\frac{1}{k})$$$${\mathrm{S}}_n=\frac{1}{2}(1+\frac{1}{2}-\frac{1}{n}-\frac{1}{n+1})$$$$\text{Double subscripts: use braces to clarify}$$

Commented by mr W last updated on 16/Sep/20

$$youareright!$$

Answered by Olaf last updated on 16/Sep/20

$$\mathrm{for}n=1,\frac{1}{n^2-1}\mathrm{is}\mathrm{not}\mathrm{defined}!$$

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