<s>[INST] Convergence test for the series $\sum_{n=1}^{\infty} \frac{\sqrt{n^2+1}-n}{\sqrt{n}}$ 
Determine convergence of  the series
  $$\sum_{n=1}^{\infty} \frac{\sqrt{n^2+1}-n}{\sqrt{n}}$$

My proof: using comparison test  I  have
$$\frac{\sqrt{n^2+1}-n}{\sqrt{n}}  =  \sqrt{n+\frac{1}{n}} - \sqrt{n}  \to 0$$ for large $n$.
Therefore the series should be converges. Now that we need to show that $a_n \lt b_n$ for some $a_n$ s.t $\sum_{n=1}^{\infty} a_n$ converges. for all $n \in\mathbb N$,
$\frac{\sqrt{n^2+1}-n}{\sqrt{n}} \lt \frac{1}{n^2}$ since $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges then $\sum_{n=1}^{\infty} \frac{\sqrt{n^2+1}-n}{\sqrt{n}}$ must be converges.
Is this right or I need to fix my $\frac{1}{n^2}$?
 [/INST] Since $$\sqrt {n^2 + 1} < n + \frac {1} {2n}$$ we have $$\frac {\sqrt {n^2 + 1} - n} {\sqrt n} < \frac {1} {2n \sqrt n}.$$ So the series is dominated by $$\sum_{n = 1}^{\infty} \frac {1} {2n \sqrt n} = \frac {1} {2} \zeta \left(\frac {3} {2} \right).$$
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