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?
Best Answer
This is just an observation, and I'm not sure how rigorous this is, but consider the difference between the linear terms in the factorization of the denominator of the terms of your series. For the first one, the difference between $n$ and $n+1$ is $1$, and notice the telescoping series when written out begins like this: $$ (1-1/2)+(1/2-1/3)+(1/3-1/4)+\cdots $$ Notice that the terms that vanish, namely $1/2$ and $1/3$, do so by the next term. That is, you only have to add $1$ more term for it to vanish from the sum.
Likewise, the difference of the linear terms $n-1$ and $n+1$ in the second series is $2$. Also, when written out, the telescoping sum is $$ (1-1/3)+(1/2-1/4)+(1/3-1/5)+(1/4-1/6)+\cdots $$ Now the terms that vanish, $1/3, 1/4$, do not do so until $2$ terms after they first appear. That is, $1/3$ first appears in term $a_1$, and doesn't disappear until term $a_3$, so you must add $2$ terms until it goes away.
The same applies for the third series, that you must add $3$ terms until the terms which will not contribute to the partial sum vanish. I believe this is helpful in figuring out how many terms you should sum before deciding on a general formula for the partial sum. Then of course induction would be the best way to make sure of your guess. I think it is just coincidence that the difference of the linear factors of the denominator was equal to the numerator in all three of the problems you listed.