I'm trying to figure out this problem and have no clue how to do it. The problem reads:
Use the identity 1/(k^2-1) = (1/2)(1/(k-1) - 1/(k+1)) to evaluate sum k=2 to n 1/(k^2-1)
the problem is almost solved, now, we have $\displaystyle \frac{1}{k-1}-\frac{1}{k+1}=\left( \frac{1}{k-1}-\frac{1}{k} \right)+\left( \frac{1}{k}-\frac{1}{k+1} \right).$ Then, if you consider $\displaystyle
\sum\limits_{k=2}^{n}{\left( \frac{1}{k-1}-\frac{1}{k} \right)},$ does it telescope? what 'bout the other one?
What he means is that a telescoping sum is a sum that can be simplified by the cancelleling of terms through the addition that it implies.
e.g.
$\displaystyle \sum_{i=1}^n(a_n-a_{n-1})$
If you look carefully at this sum you can see that no matter how large n is that there will only be two terms left standing after what I call the addition wars. check it out
$\displaystyle \sum_{i=1}^n(a_n-a_{n-1})$
$\displaystyle =(a_n-a_{n-1})+(a_{n-1}-a_{n-2})+(a_{n-2}-a_{n-3})+...+(a_{3}-a_{2})+(a_{2}-a_{1})$
So after an uncountable number of n terms, who comes out of this war alive?
$\displaystyle \sum_{i=1}^n(a_n-a_{n-1})=a_{n}-a_{1}$
So now look at your problem with a trained eye. You have the ability to predict who's side you shoulb be on!