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Investigate convergence of the series: $$\frac{1}{n( \sqrt{n^2+n}-n)}$$ Which criterion should be used?

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  • $\begingroup$ As a sequence or a series? $\endgroup$ – Tim Raczkowski Feb 18 '15 at 14:21
  • $\begingroup$ series, my mistake $\endgroup$ – kurkowski Feb 18 '15 at 14:22
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Or rewrite:

$$\frac{1}{n\left( \sqrt{n^2+n}-n \right)} = \frac{\sqrt{n^2+n}+n}{n\left( \sqrt{n^2+n}-n \right)\left( \sqrt{n^2+n}+n \right)} = \frac{\sqrt{n^2+n}+n}{n^2} $$

And then:

$$\frac{\sqrt{n^2+n}+n}{n^2} \ge \frac{\sqrt{n^2}+n}{n^2} = \frac{2n}{n^2} = \frac{2}{n} $$

So since $\sum \tfrac{2}{n}$ diverges...

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  • $\begingroup$ You're welcome! $\endgroup$ – StackTD Feb 18 '15 at 14:43
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Hint. You may write, as $n \to +\infty$: $$ \frac{1}{n( \sqrt{n^2+n}-n)}=\frac{1}{n^2( \sqrt{1+\frac1n}-1)}=\frac{1}{n^2\left(\frac{1}{2n}+\mathcal{O}\left({\frac1{n^2}}\right)\right)}\sim \frac2n $$ and the series $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n( \sqrt{n^2+n}-n)}$ is divergent.

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  • $\begingroup$ It seems only French users systematically resort to $\sim$ asymptotical evaluations to prove convergence/divergence of series. Most people are limited to comparison test, which is too bad. $\endgroup$ – Gabriel Romon Feb 18 '15 at 15:55
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Right answer is: $$ \frac{1}{n \cdot (\sqrt{n^2+n} - n) } = \frac{(\sqrt{n^2+n} + n)}{n \cdot (\sqrt{n^2+n} - n) \cdot (\sqrt{n^2+n} + n) } = \frac{n \cdot (\sqrt{1+\frac{1}{n}} + 1)}{n \cdot (n ^2 + n - n^2)} = \frac{\sqrt{1+\frac{1}{n}} + 1}{n } $$ And then you can do with this whatever you want. This is quite common technic for such problems.

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