$$\sum_{n=2}^\infty\frac1{n^3-n}=?$$
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1Hint: $n^3-n=(n-1)n(n+1)$ – Parcly Taxel Feb 10 '17 at 06:36
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1Hint: Use the comparison test to see if the series converges. – Harnoor Lal Feb 10 '17 at 06:39
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See http://math.stackexchange.com/questions/2110348/find-the-sum-of-infinite-series-frac12-cdot-3-cdot-4-frac14-cdot-5-cd – lab bhattacharjee Feb 10 '17 at 06:50
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If you can derive an expression for the partial sum, then you can determine what the series converges to. – SOULed_Outt Feb 10 '17 at 06:53
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HINT : $$\sum_{n=2}^\infty\frac1{n^3-n}=\frac{1}{2}\sum_{n=2}^\infty\left(\frac1{n+1}+\frac1{n-1}-\frac2{n}\right)=\frac{1}{4}\quad\text{because telescoping series.}$$
JJacquelin
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Write down the series without the $\sum$. You will see that almost all terms canceled one to another. Only a few terms are remaining which sum is 1/4. That is what is called "telescoping series". – JJacquelin Feb 10 '17 at 09:33
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Take care of the comment from Harnoor Lal and edit your question to show that you worked out the prove of convergence. It is necessary before any further calculus. If you don't edit your work, your question will probably be deleted. – JJacquelin Feb 10 '17 at 09:45