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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?

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hm, I just came upon a problem that doesn't involve fractions, but e. I guess it was pure coincidence. What determines, then, how far I must go? – ShrimpCrackers Jan 13 '11 at 6:24
What was the question? is it related to telescoping series? – Arjang Jan 13 '11 at 15:10
@Arjang: You can find an exponential like telescoping series in the third line of my answer to this question:… – Derek Jennings Jan 13 '11 at 15:33
up vote 4 down vote accepted

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.

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Thanks yunone. I don't think I would have spotted this on my own. – ShrimpCrackers Jan 13 '11 at 16:28

It is just a coincidence that the number of terms to keep equals the numerator. In your second example, if your were summing $\frac{1}{n^2-1}$ you would still keep two terms.

All your examples are of the form $\frac{a}{n(n+a)}$ (in the second you need to shift $n$ by 1) and the number of terms to keep is $a$. This is because $\frac{a}{n(n+a)}=\frac{1}{n}-\frac{1}{n+a}$ so terms will telescope $a$ spaces down the line. The number of terms to keep depends upon the denominator, not the numerator.

You were given $a$ in the numerator to avoid a fraction when making this transformation.

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If your terms are hypergeometric then you can use Gosper's algorithm.

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