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