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!
 A: We have
$$ \frac1{2\sqrt{n+1}}< \frac{1}{\sqrt{n+1}+\sqrt{n}} < \frac1{2\sqrt n} \tag 1$$
As $$\sum_{n=1}^\infty \frac1 {n^y} \quad\text{ converges iff }\quad y>1 \tag2$$
it follows that
$$\sum_{n=1}^\infty \frac1 {(\sqrt{n+1}+\sqrt{n})^\alpha}$$
converges iff $\alpha > 2$. Thus, it converges in particular for $\alpha\geqslant 4$.
A: Use the limit comparison test
Consider :-
$$\lim_{n\to\infty}\frac{\Bigg(\frac{1}{\sqrt{n+1}+\sqrt{n}}\Bigg)^{\alpha}}{\frac{1}{n^p}}=\lim_{n\to\infty}\frac{\left(\frac{1}{\sqrt{n+1}}\right)^{\alpha}\Bigg(\frac{1}{1+\sqrt{\frac{n}{n+1}}}\Bigg)^{\alpha}}{\frac{1}{n^p}}=\lim_{n\to\infty}\frac{\left(\frac{1}{\sqrt{n+1}}\right)^{\alpha}}{\frac{1}{n^{p}}}\cdot\frac{1}{2^{\alpha}}$$
Then this limit is $\frac{1}{4}$ for $\alpha=2$ and $p=1$ hence it diverges as $\sum\frac{1}{n}$ diverges
for $\alpha>2$ and $p=\frac{\alpha}{2}$ the limit is again finite and positive and as $p>1$ the series converges as $\sum\frac{1}{n^{p}}$ converges for $p>1$
And for $\alpha<2$ and $p=\frac{\alpha}{2}$ the limit is again finite and as $p<1$ it diverges as $\sum\frac{1}{n^{p}}$ diverges for $p\leq 1$.
A: 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 $\alpha > 1$.
