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$$\sum_{n=2}^\infty\frac1{n^3-n}=?$$

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    _Hint_: $n^3-n=(n-1)n(n+1)$ – 2017-02-10
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    Hint: Use the comparison test to see if the series converges. – 2017-02-10
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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 – 2017-02-10
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    If you can derive an expression for the partial sum, then you can determine what the series converges to. – 2017-02-10

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HINT : $$\sum_{n=2}^\infty\frac1{n^3-n}=\frac{1}{2}\sum_{n=2}^\infty\left(\frac1{n+1}+\frac1{n-1}-\frac2{n}\right)=\frac{1}{4}\quad\text{because telescoping series.}$$

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    Could you please tell me how you found the limit 1\4 – 2017-02-10
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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". – 2017-02-10
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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. – 2017-02-10