# Determine the sum of the Telescopic serie

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 $$\sum_{n=2}^\infty\frac{2}{n^2-1}$$

• Why don't u try with $$\dfrac2{n^2-1}=\dfrac{n+1-(n-1)}{(n+1)(n-1)}=f(n-1)-f(n+1)$$ where $$f(m)=\dfrac1m$$ – lab bhattacharjee Jun 7 '17 at 3:23

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)?

• I did the partial fraction decomposition, but when i opened term by term i can't see the result – Hamilton Junior Jun 7 '17 at 3:27
• @HamiltonJunior see my addendum – Dave Jun 7 '17 at 3:34
• now i saw and two terms left if i correct. $1$ and $\frac{1}{2}$, Correct? – Hamilton Junior Jun 7 '17 at 3:37
• Those are not quite the only terms left in the $N$-th partial sum. However, after taking the limit to infinity these are the remaining terms. Can you see what the other two terms remaining in the $N$-th partial sum should be? – Dave Jun 7 '17 at 3:38
• $-\frac{1}{N}$ and $\frac{1}{N-3}$ ? – Hamilton Junior Jun 7 '17 at 3:42

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

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

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

Hint: $$\int_0^1x^{n-2}-x^ndx=\frac{2}{n^2-1}$$ using this hint we will get $$\sum_{n=2}^{\infty}{\frac{2}{n^2-1}}=\int_0^1{\sum_{n=2}^{\infty}x^{n-2}-\sum_{n=2}^{\infty}x^ndx}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int_0^1{\sum_{n=0}^{\infty}x^{n}-\sum_{n=2}^{\infty}x^ndx}\\ \ \ \ =\int_0^1{(1+x)dx}$$This is an integral one can do via standard calculus methods.