Calculate:
$\sum \limits_{n=2}^{\infty}\dfrac{1}{n^3-n}$
I have tried to find the sequence partial sums and show that they converge, but I'm having trouble setting it up and I have no idea how to calcuate the numerical value.
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2comparison test. for n big enough n^3 - n > n^2 – J. Dionisio Feb 18 '17 at 11:55
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1Related : http://math.stackexchange.com/questions/2110348/find-the-sum-of-infinite-series-frac12-cdot-3-cdot-4-frac14-cdot-5-cd – lab bhattacharjee Feb 18 '17 at 12:01
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For convergence, note that
$$n^2<n^3-n\implies\frac1{n^3-n}<\frac1{n^2}$$
Which holds true for $n\ge1$.
To evaluate, notice that $n^3-n=n(n-1)(n+1)$, and thus we have a nice PFD:
$$\frac1{n^3-n}=\frac12\left[\frac1{n(n-1)}-\frac1{n(n+1)}\right]$$
Which gives a telescoping series, so,
$$\sum_{n=2}^\infty\frac1{n^3-n}=\frac14$$
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Note that this is wrong by a factor of $1/2$, compare e.g. https://math.stackexchange.com/a/1663085/42969. – Martin R Apr 04 '18 at 20:10
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