<s>[INST] convergence of the series $\sum_{n=1}^\infty \frac {\sqrt{n^2+1}-n}{\sqrt{n}}$ 
I'm studying the convergence of the series
  $$\sum_{n=1}^\infty \frac {\sqrt{n^2+1}-n}{\sqrt{n}}$$



*

*$\frac {\sqrt{n^2+1}-n}{\sqrt{n}}>0, \forall n \ge 1$

*$\lim_{n \to +\infty}\frac {\sqrt{n^2+1}-n}{\sqrt{n}}=\lim_{n \to +\infty}\frac {n^2+1-n^2}{\sqrt{n}*(\sqrt{n^2+1}+n)}=0$

*The suggested solution in my book says that the series converges but:
$\frac {\frac {\sqrt{n^2+1}-n}{\sqrt{n}}}{(\frac {1}{n})^{- \frac {1}{2}}}= \frac {n* \sqrt {1+ \frac {1}{n^2}}}{ \sqrt {n}}* \frac {1}{\sqrt {n}}    \rightarrow 1 $ for $n \rightarrow \infty $

 [/INST] Note that
$$ \frac {\sqrt{n^2+1}-n}{\sqrt{n}} \cdot \frac { \sqrt{n^2+1}+n }{\sqrt{n^2+1}+n}= \frac { 1 }{\sqrt{n^3+n}+n\sqrt n}\sim \frac{1}{2\sqrt{n^3}}$$
thus
$$\sum_{n=1}^\infty \frac {\sqrt{n^2+1}-n}{\sqrt{n}}=\sum_{n=1}^\infty \frac { 1 }{\sqrt{n^3+n}+n\sqrt n}$$
converges by comparison test with $$\frac{1}{\sqrt{n^3}}$$
</s>