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Find:

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

I tried to resolve into partial fractions to see if there are some cancellations ,but that did not helped me .How do i do this ?Thanks

Thomas Andrews
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godonichia
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  • It is an infinite series so when you say limit as $n$ goes to infinity, it does not make any sense. Try partial fractions $\frac{1}{(n-1)n(n+1)} = \frac{1}{2}\left(\frac{1}{n-1}+\frac{1}{n+1} - \frac{2}{n}\right)$ .. there will be cancellation. – r9m Feb 09 '15 at 04:04
  • i tried to do this ,but calculations didnot seem to happen – godonichia Feb 09 '15 at 04:04
  • The $N^{th}$ partial sum $\sum\limits_{n=2}^{N} \frac{1}{n^3-n} = \sum\limits_{n=2}^{N} \frac{1}{2}\left(\frac{1}{n-1}+\frac{1}{n+1} - \frac{2}{n}\right) =\frac{1}{2} \sum\limits_{n=2}^{N} \frac{1}{n-1}+\frac{1}{2}\sum\limits_{n=2}^{N} \frac{1}{n+1}-\sum\limits_{n=2}^{N} \frac{1}{n}$ .. these $3$ summations do have common terms and they will cancel. – r9m Feb 09 '15 at 04:08

2 Answers2

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Since $n^3 - n = n(n^2 - 1) = n(n-1)(n+1)$, $$\frac{1}{n^3 - n} = \frac{1}{n}\frac{1}{(n-1)(n+1)} = \frac{1}{2n}\left(\frac{1}{n-1} - \frac{1}{n+1}\right) = \frac{1}{2(n-1)n} - \frac{1}{2n(n+1)}.$$ Now $\sum_{n = 2}^\infty 1/(n^3 - n)$ telescopes to $1/4$.

kobe
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We know that the series at hand is convergent absolutely, so we can safely play with the sum.

$$\sum_{i=2}^\infty\frac{1}{n^3 - n} = -\sum_{i=2}^\infty\frac{1}{n} + 0.5\sum_{i=2}^\infty\frac{1}{n+1} + 0.5\sum_{i=2}^\infty\frac{1}{n-1} = -\sum_{i=2}^\infty\frac{1}{n} + 0.5\sum_{i=3}^\infty\frac{1}{n} + 0.5\sum_{i=1}^\infty\frac{1}{n} = -\sum_{i=1}^\infty\frac{1}{n} + 1 + 0.5\sum_{i=1}^\infty\frac{1}{n} - 0.5 - 0.25 + 0.5\sum_{i=1}^\infty\frac{1}{n} = 1 - 0.75 = 0.25 = \frac{1}{4}$$.

As $n \rightarrow \infty$, the limit is $\frac14$.