# Convergence of $\frac{1}{n( \sqrt{n^2+n}-n)}$

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

• As a sequence or a series? – Tim Raczkowski Feb 18 '15 at 14:21
• series, my mistake – kurkowski Feb 18 '15 at 14:22

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...

• You're welcome! – StackTD Feb 18 '15 at 14:43

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.

• 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. – Gabriel Romon Feb 18 '15 at 15:55

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.