I'm confused as to how far to go in terms of getting $s_n$ in a telescoping series. I cannot find an explanation for this at all in my textbook. But from the problems I've seen, perhaps it's just coincidence, does it have anything to do with the numerator?
For example, the section example gives:
$(\sum_{n=1}^\infty) \frac{1}{n(n+1)}$
$s_n = 1 - \frac{1}{n+1}$
In another problem I've noticed:
$(\sum_{n=2}^\infty) \frac{2}{n^2-1}$
$s_n = 1 + \frac{1}{2} - \frac{1}{n-1} - \frac{1}{n}$
In another problem:
$(\sum_{n=1}^\infty) \frac{3}{n(n+3)}$
$s_n = 1 + \frac{1}{2} + \frac{1}{3} - \frac{1}{1+n} - \frac{1}{2+n} - \frac{1}{3+n}$
The fractions with denominator n become 0, so they don't count, but I noticed that in the collapsing series, the fractions that did matter equaled the numerator. Just a coincidence? If not, why does this happen?