How to find $\sum_{n=2}^{+\infty}\frac{1}{(n^2-1)}$

I'm completely lost. Wolfram alpha says that partial sums look like this: $$\sum_{n=2}^{+\infty}\frac{1}{(n^2-1)}=\frac{3m^2-m-2}{4m(m+1)}$$ From here, the sum is clearly equal to $\frac34$. But how on earth does one come up with such formula? Maybe I'm missing something obvious here, but I can't figure it out. Thanks!

• Use partial fractions and you will as lab has said its a telescoping series. – user60887 Dec 5 '13 at 17:37
• I think you can also get this from a fourier series of $\cos x$ or $\cosh x$ or something to that effect, but I can't recall. – abnry Dec 5 '13 at 17:43

HINT:

If $n^2\ne1\iff n\ne\pm1,$

$$\frac1{n^2-1}=\frac12\frac{(n+1)-(n-1)}{(n-1)(n+1)}=\frac12\left(\frac1{n-1}-\frac1{n+1}\right)$$

Can you recognize the Telescoping series ?

• Good hint...you beat me to it. – user113529 Dec 5 '13 at 17:40
• Interesting, we didn't have it on our lecture... Thanks for the link, it's very helpful! – Dunno Dec 5 '13 at 17:44
• – lab bhattacharjee Dec 5 '13 at 17:50
• – lab bhattacharjee Dec 5 '13 at 17:51
• haha wow, thanks! It's always good to know I'm not the only one who doesn't see something. – Dunno Dec 5 '13 at 17:52

I'll give you a hint. Factor the bottom to get $$\frac{1}{(n+1)(n-1)}$$ Now use partial fractions to write $$\frac{1}{(n+1)(n-1)}=\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)$$ Can you compute the $M$th partial sum rewriting in this way? Look for cancellations.

• One small comment...just rememeber to calculate the limit as $M\to \infty$ when your done computing the $M$th partial sum. You will usually get a number $+$ something that goes to $0$, but it is possible to have a situation where you have a number $+$ a nonzero limit. – user113529 Dec 5 '13 at 17:43

The formula you obtained is for a summation from 2 to m and it is correct. If "m" goes to an infinite value, the limit is 3/4. Could you check what you submitted to WA ? Thanks for letting me know.