Determine all $q \in \mathbb{Q}$, so that $ \sum_{n=1}^{\infty}{\frac{\sqrt{n+1}-\sqrt{n}}{n^q}} $ converges

Question

I already tried the ratio and root criterion, but it didn't get me anywhere. I'd be thrilled if you had any suggestions.

(also applied the third binomial formula, so it would "look nicer", maybe it helps $\sum_{n=1}^{\infty}{\frac{\sqrt{n+1}-\sqrt{n}}{n^q}}=\sum_{n=1}^{\infty}{\frac{1}{n^q(\sqrt{n+1}+\sqrt{n})}}$)

thanks and greetings!

Answer 1

Clearly $$ \frac{1}{2(n+1)^{q+{1/2}}}<\frac{\sqrt{n+1}-\sqrt{n}}{n^q}=\frac{1}{(\sqrt{n+1}+\sqrt{n})\cdot n^q}<\frac{1}{2n^{q+{1/2}}}, $$ and $\sum_{n=1}^\infty\frac{1}{n^{q+1/2}}$, equivalently $\sum_{n=1}^\infty\frac{1}{(n+1)^{q+1/2}}$, converge, if and only if $q+1/2>1$, equivalently $q>1/2$.

Answer 2

From here

$$\sum_{n=1}^{\infty}{\frac{1}{n^q(\sqrt{n+1}+\sqrt{n})}}$$

we have that

$$\frac{1}{n^q(\sqrt{n+1}+\sqrt{n})}\sim \frac{1}{2n^{q+\frac12}}$$

then refer to limit comparison test to conlcude that $q+\frac12>1$.

Answer 3

Hint 1:

$${\frac{\sqrt{n+1}-\sqrt{n}}{n^q}}={\frac{1}{n^q(\sqrt{n+1}+\sqrt{n})}}$$

Hint 2 Limit compare ${\frac{1}{n^q(\sqrt{n+1}+\sqrt{n})}}$ to ${\frac{1}{n^q\sqrt{n}}}$