Determine the sum of the Telescopic serie

Question

I Can't see where the serie goes.

I opened N terms, but I'm having some issues. If someone can help. The serie is $$\sum_{n=2}^\infty\frac{2}{n^2-1}$$

Answer 1

Hint: $$\frac{2}{n^2-1}=\frac{2}{(n-1)(n+1)}=\frac{1}{n-1}-\frac{1}{n+1}$$ Can you see how this is a telescopic sum?

Addendum: Looking at the $N$-th partial sum we have $$\begin{align}\sum_{n=2}^N\left(\frac{1}{n-1}-\frac{1}{n+1}\right)&=\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{4}-\frac{1}{6}\right)+...+\\&\ \ \ \ \ \left(\frac{1}{N-3}-\frac{1}{N-1}\right)+\left(\frac{1}{N-2}-\frac{1}{N}\right)+\left(\frac{1}{N-1}-\frac{1}{N+1}\right)\\&=1+\frac{1}{2}+\left(-\frac{1}{3}+\frac{1}{3}\right)+\left(-\frac{1}{4}+\frac{1}{4}\right)+...\end{align}$$ Do you see which terms will cancel out (Hint: there should be two terms left over on each end of the sum)?

Answer 2

$S_k=\sum_{n=2}^k\{\frac{1}{n-1}-\frac{1}{n+1}\}$

$=1+\frac{1}{2}-\frac{1}{k}-\frac{1}{k+1}$

Hence series converges to $\frac{3}{2}$