convergence of the series $\sum_{n=1}^\infty \frac {\sqrt{n^2+1}-n}{\sqrt{n}}$

Question

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 $

Answer 1

Now, $$\frac{1}{\sqrt{n}(\sqrt{n^2+1}+n)}<\frac{1}{2\sqrt{n^3}}$$ and since $\frac{3}{2}>1$ it converges.

Because $$\sum_{n=1}^{+\infty}\frac{\sqrt{n^2+1}-n}{\sqrt{n}}<\sum_{n=1}^{+\infty}\frac{1}{2\sqrt{n^3}}$$ and since $\sum\limits_{n=1}^{+\infty}\frac{1}{2\sqrt{n^3}}$ converges, we are done.

Answer 2

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 $$\frac{1}{\sqrt{n^3}}$$