The convergence of $\sum ( \sqrt{n+1}-\sqrt{n})^ a$. How do I keep it from here? Check when the series $\sum ( \sqrt{n+1}-\sqrt{n})^ a$, $a\in \Bbb{R}$ converges.

$Attempt$: Let us focus on $a_n= ( \sqrt{n+1}-\sqrt{n})^ a$. $a_n=( \sqrt{n+1}-\sqrt{n})^ a=( (\sqrt{n+1}-\sqrt{n})\cdot{\sqrt{n+1}+\sqrt{n}\over \sqrt{n+1}+\sqrt{n}})^ a=({1\over \sqrt{n+1}+\sqrt{n}})^a$. 

I know I sort of have to find an $a$ for which $({1\over \sqrt{n+1}+\sqrt{n}})^a\le {1\over n^b}$ where $b&gt;1$. Any suggestion for a way to $resume$ my attempt? Thank you.
Okay it remained unsolved but I solved it so if it means anything: 
$({1\over \sqrt{n+1}+\sqrt{n}})^a\le ({1\over 2\sqrt{n}})^a$.

If $\sum ({1\over 2\sqrt{n}})^a$ converges so does the original one. 

Therefore, $\sum ({1\over 2\sqrt{n}})^a=({1\over 2})^a\sum ({1\over\sqrt{n}})^a$ converges $\iff$ $\sum ({1\over\sqrt{n}})^a$ converges. 

But, $\sum ({1\over\sqrt{n}})^a=\sum ({1\over\ n^{1\over 2}})^a=\sum {1\over\ n^{a\over 2}}$ 
By $p-series$ test, $\sum {1\over\ n^{a\over 2}}$ converges $\iff$ $a&gt;2.$
$$ \frac{1}{\sqrt{n+1}+\sqrt{n}}&lt;\frac{1}{\sqrt{n}+\sqrt{n}}=\frac{1}{2\sqrt{n}}\\so\\(\frac{1}{2\sqrt{n}})^a\leq(\frac{1}{2\sqrt{n}})^b\\$$