Test the convergence of the series... So, the series is:
$$\sum_{n=1}^\infty\frac{\sqrt{n+2}-\sqrt{n}}{n^k}$$
What I did is:
$$a_n=\frac{\sqrt{n+2}-\sqrt{n}}{n^k}*\frac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+2}+\sqrt{n}}=\frac{2}{n^k(\sqrt{n+2}+\sqrt{n})}$$
I'm not sure what to do next, which criteria should I use?

Note
$$\frac{1}{2n^{k+1/2}} < a_n < \frac{2}{n^{k+1/2}}$$
for all $n \ge 1$. Thus $\sum\limits_{n = 1}^\infty a_n$ converges if and only if $\sum\limits_{n = 1}^\infty n^{-(k+1/2)}$ converges. Now use the $p$-test.
Remark: To get the lower bound $\frac{1}{2n^{k + 1/2}} < a_n$, consider that 
$$\sqrt{n+2} + \sqrt{n} <2\sqrt{n+2} < 2\sqrt{n + 3n} = 2\sqrt{4n} = 4\sqrt{n}.$$
This implies $a_n > \frac{2}{n^k(4\sqrt{n})} = \frac{1}{2n^{k+1/2}}$.
