Telescoping Summation problem!

$Notice\: that,\: \frac{1}{n^{2}-1}=\frac{1}{2} \times \left ( \frac{1}{n-1}-\frac{1}{n+1} \right )$

$Now,$ $\sum_{n=2}^{\infty} \frac{1}{n^{2}-1}=\lim_{k \to \infty}\sum_{n=2}^{k}\frac{1}{n^{2}-1}$ $=\frac{1}{2} \times \lim_{k \to \infty}\sum_{n=2}^{k}\left ( \frac{1}{n-1}-\frac{1}{n+1} \right )$

$Now,\: notice\: the\: term,\: \lim_{k \to \infty}\sum_{n=2}^{k}\left ( \frac{1}{n-1}-\frac{1}{n+1} \right )$ $=\lim_{k \to \infty} \left [ \left ( \frac{1}{1}-\frac{1}{3} \right )+\left ( \frac{1}{2}-\frac{1}{4} \right )+\left ( \frac{1}{3}-\frac{1}{5} \right )+...+\left ( \frac{1}{k-2}-\frac{1}{k} \right )+\left ( \frac{1}{k-1}-\frac{1}{k+1} \right ) \right ]$ $=\lim_{k \to \infty} \left [ \frac{1}{1} + \frac{1}{2} - \frac{1}{k} - \frac{1}{k+1}\right ] \left [ Because\:each\:term\:telescopes\:without\:the\:following\:four \right ]$ $=\frac{1}{1} + \frac{1}{2} - \lim_{k \to \infty} \left [ \frac{1}{k} +\frac{1}{k+1}\right ]$ $=\frac{3}{2}+0$

$Thus\: we\: get,\: \frac{1}{2} \times \lim_{k \to \infty}\sum_{n=2}^{k}\left ( \frac{1}{n-1}-\frac{1}{n+1} \right ) = \frac{1}{2} \times \frac{3}{2}$ $=\frac{3}{4}$ $=\boxed{0.75}$

Actually the $\lim _{k \rightarrow \infty}$ is unnecessary.

$\sum _{n=2} ^{\infty} {\frac {1}{n^2-1} } = \frac {1}{2} \sum _{n=2} ^{\infty} { \left( \frac {1}{n-1} - \frac {1}{n+1} \right)}$

$= \frac {1}{2} \left( \frac {1}{1} - \frac {1}{3} + \frac {1}{2} - \frac {1}{4} + \frac {1}{3} - \frac {1}{5} + \frac {1}{4} - \frac {1}{6} + \frac {1}{5} - \frac {1}{7} + ... \right)$

It can be seen that:

• the $2^{nd}$ term $-\frac {1}{3}$ cancels with $5^{th}$ term $+\frac {1}{3}$
• the $4^{th}$ term $-\frac {1}{4}$ cancels with $7^{th}$ term $+\frac {1}{4}$
• the $6^{th}$ term $-\frac {1}{5}$ cancels with $9^{th}$ term $+\frac {1}{5}$
• and so on until $\infty$ .

Therefore,

$\sum _{n=2} ^{\infty} {\frac {1}{n^2-1} } = \frac {1}{2} \left( 1 + \frac {1}{2} \right) = \frac {1}{2} \times \frac {3}{2} = \frac {3}{4} = \boxed {0.75}$

Find an expression so that the variables (fractions) will cancel out each other.

$\sum _{ n=2 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 }-1 } } \\ =\frac { 1 }{ { 2 }^{ 2 }-1 } +\frac { 1 }{ { 3 }^{ 2 }-1 } +...+\frac { 1 }{ { n }^{ 2 }-1 }$

We find out that

$\frac { 1 }{ { 2 }^{ 2 }-1 } =\frac { 1 }{ 2 } (1-\frac { 1 }{ 3 } )\\ \frac { 1 }{ { 3 }^{ 2 }-1 } =\frac { 1 }{ 2 } (\frac { 1 }{ 2 } -\frac { 1 }{ 4 } )\\ \frac { 1 }{ { 4 }^{ 2 }-1 } =\frac { 1 }{ 2 } (\frac { 1 }{ 3 } -\frac { 1 }{ 5 } )\\ \frac { 1 }{ { 5 }^{ 2 }-1 } =\frac { 1 }{ 2 } (\frac { 1 }{ 4 } -\frac { 1 }{ 6 } )\\ \qquad \qquad ...\\ \frac { 1 }{ { n }^{ 2 }-1 } =\frac { 1 }{ 2 } (\frac { 1 }{ n-1 } -\frac { 1 }{ n+1 } )$

And the variables after $\frac { 1 }{ 2 } (1)\quad$ and $\frac { 1 }{ 2 } (\frac { 1 }{ 3 } )$ cancel out each other

Since $\frac { 1 }{ n+1 }$ has a negligible value,

$\sum _{ n=2 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 }-1 } } \\ =\frac { 1 }{ 2 } (1+\frac { 1 }{ 2 } )\\ =0.75$