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 $b>1$. Any suggestion for a way to $resume$ my attempt? Thank you. 
 A: 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 $\iff$ $a>2.$
A: $$\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n}+\sqrt{n+1}}\in\left(\frac{1}{2\sqrt{n+1}},\frac{1}{2\sqrt{n}}\right),$$
so $(\sqrt{n+1}-\sqrt{n})^a$ behaves like $\color{red}{(4n)^{-a/2}}$.
A: I'm sorry I'm posting this as an answer, but this is solely because I can't leave the comment. This one is related to the @Meitar's self-answer.
@Meitar, you only half-solved it, because you showed that original series converge when $a>2$, but it doesn't follow from your solution that it diverges when $a \leq 2$.
To make it clear, you need to do the following. One have to say that if $\sum ({1\over 2\sqrt{n+1}})^a$ diverges so does the original one, because $({1\over \sqrt{n+1}+\sqrt{n}})^a \ge ({1\over 2\sqrt{n+1}})^a$. Repeating your proof for convergence we conclude that original series diverges when diverges $\sum {1\over\ (n+1)^{a\over 2}}$. The latter happens when $a \leq 2$, because $+1$ of course doesn't matter.
P.S. I need +5 rep to be able to leave comments. If you found this useful please upvote so that I can put comments in the future instead of spamming the answer :( Thanks!
A: $$ \frac{1}{\sqrt{n+1}+\sqrt{n}}<\frac{1}{\sqrt{n}+\sqrt{n}}=\frac{1}{2\sqrt{n}}\\so\\(\frac{1}{2\sqrt{n}})^a\leq(\frac{1}{2\sqrt{n}})^b\\$$
