Prove that for $\alpha \leq2$ the sum $\sum_{n=1}^\infty (\sqrt{n+1}-\sqrt{n})^\alpha$ is divergent, and if $\alpha \geq 4$ the sum is convergent. Prove that for $\alpha \leq2$ the sum $\sum_{n=1}^\infty (\sqrt{n+1}-\sqrt{n})^\alpha$ is divergent, and if $\alpha \geq 4$ the sum is convergent.
Attempt:
$$\sum_{n=1}^\infty (\sqrt{n+1}-\sqrt{n})^\alpha = \sum_{n=1}^\infty ((\sqrt{n+1}-\sqrt{n})\cdot\dfrac{(\sqrt{n+1}+\sqrt{n})}{(\sqrt{n+1}+\sqrt{n})})^\alpha =$$
$$\sum_{n=1}^\infty (\dfrac{1}{\sqrt{n+1}+\sqrt{n}})^\alpha$$
Therefore, we will show for every $\alpha \leq 2$ these will going to make the sum diverge.
Easy to see that when $\alpha < 0 $ we get that:
$$\lim_{n\rightarrow \infty}(\dfrac{1}{\sqrt{n+1}+\sqrt{n}})^\alpha \neq 0$$
If $\alpha =1 $ then we get:
$$\sum_{n=1}^\infty (\dfrac{1}{\sqrt{n+1}+\sqrt{n}})$$
So, we can see that the sum is always positive so we can use the limit test of the sums,
Let $b_n = \sum_{n=1}^\infty (\dfrac{1}{\sqrt{n+1}+\sqrt{n}})$ and let $a_n = \sum_{n=1}^\infty (\dfrac{1}{\sqrt{n}})$
And we get $\lim_{n\rightarrow \infty}(\dfrac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n}}) =\lim_{n\rightarrow \infty}(\dfrac{\sqrt{n+1}}{\sqrt{n}}+1)=$ $\lim_{n\rightarrow \infty}(\sqrt{1+\dfrac{1}{n}}+1) = 2$
Therefore it diverges too.
For $\alpha=2$, I get struggled with that and I didn't succeed in proving it and how I can prove that converges for all $\alpha \geq 4$.
Thanks!

You could write $\sqrt{n+1} - \sqrt{n} = \sqrt{n}\left(\sqrt{1 + \frac{1}{n}} - 1\right)$ and then note that for $n\geqslant 1$,
$$
\frac{\sqrt 2 -1}{n} \leqslant \left( \sqrt{1+\frac{1}{n}} -1 \right) \leqslant\frac{1}{n}
$$
where the right hand inequality is derived from the basic property $x \geq \sqrt x$ when $x \geqslant 1$ and the left inequality from $\sqrt{x} \geqslant 1 + (\sqrt2 - 1)(x-1)$ when $1 \leqslant x \leqslant 2$ (picture the graph of $\sqrt x$ and its chord between $x=1$ and $x=2$).
Then,
$$
\frac{(\sqrt 2- 1 )^\alpha}{n^{\alpha/2}} \leqslant (\sqrt{n+1}-\sqrt{n})^\alpha \leqslant \frac{1}{n^{\alpha/2}}.
$$
Thus your sum converges if and only if $\sum n^{-\alpha/2}$ converges, which is the case if and only if $\alpha > 1$.
