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Mathematics 10 Online
OpenStudy (anonymous):

sum_{n=2}^{infty} [2/(n^2-1)] with steps

OpenStudy (anonymous):

\[\sum_{n=2}^{\infty} [2/(n^2-1)]\]

OpenStudy (anonymous):

rewrite as \[\frac{2}{n^2-1}=\frac{1}{n-1}-\frac{1}{n+1}\] the see that you have a telescoping sum of which only the first two terms survive

OpenStudy (anonymous):

adding up leaves \[1+\frac{1}{2}=\frac{3}{2}\]

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