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How do you determine if a telescoping series is convergent or not? If it converges, what value does it converge to?

It seems like you need to do partial fraction decomposition and then evaluate each term individually?

For example:

$$ \sum_{n=2}^\infty \frac{1}{n^3-n} $$

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  • $\begingroup$ I'm not sure what you can say in general, other than "try to determine if it's telescoping." $\endgroup$ – Sammy Black Apr 5 '13 at 0:33
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    $\begingroup$ Bad example: this one is absolutely convergent. No need to worry about telescoping anything. Indeed, if $x_n=\frac{1}{n^3-n}$, then $0\leq x_{n+1}=\frac{1}{n^3+3n^2+2n+1}\leq \frac{1}{n^3}$. And the latter is the general term of a Riemann $p$ series which converges. $\endgroup$ – Julien Apr 5 '13 at 0:37
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A telescoping series is of the form $(a_1-a_2)+(a_2-a_3)+(a_3-a_4)+\ldots$. The terms in the sequence of partial sums are $a_1-a_2,a_1-a_3,a_1-a_4,\ldots$ and this converges if and only if $(a_n)$ converges.

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    $\begingroup$ In the example that he gives, the telescoping is slightly more complicated, though. $\endgroup$ – Sammy Black Apr 5 '13 at 0:59
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    $\begingroup$ @SammyBlack How so? Consider $a_n=1/(2n(n-1))$. $\endgroup$ – Did Apr 22 '13 at 14:55
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Partial fraction decomposition in your example yields $$ \frac{1}{n^3 - n} = \frac{1}{2} \left[ \frac{1}{n - 1} - \frac{2}{n} + \frac{1}{n + 1} \right] $$

To see the telescoping of the sum, it's nice to arrange terms like this:

$$ \begin{array}{*{13}{c}} 2 \sum_{n = 2}^\infty \frac{1}{n^3 - n} &=& \frac{1}{1} &-& \frac{2}{2} &+& \frac{1}{3} && && && \\ && &+& \frac{1}{2} &-& \frac{2}{3} &+& \frac{1}{4} && && \\ && && &+& \frac{1}{3} &-& \frac{2}{4} &+& \frac{1}{5} && \\ && && && &+& \frac{1}{4} &-& \frac{2}{5} &+& \frac{1}{6} \\ && && && && &+& \cdots && \end{array} $$

Now it's clear that the $N$th partial sum $(N \ge 3)$ is $$ s_N = \sum_{n = 2}^N \frac{1}{n^3 - n} = \frac{1}{2} \left[ \frac{1}{1} - \frac{1}{2} - \frac{1}{N} + \frac{1}{N + 1} \right]. $$

The infinite series is the limit of these partial sums:

$$ \sum_{n = 2}^\infty \frac{1}{n^3 - n} = \lim_{N \to \infty} s_N = \frac{1}{2}(\frac{1}{1} - \frac{1}{2}) = \frac{1}{4}. $$

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