Test for convergence: $$\sum\limits_{n=1}^\infty n^p(\sqrt{n+1}-\sqrt{n})$$ where $p\in\mathbb{R}$
My attempt:
Case 1: $p\ge0$
$\forall n\in\mathbb{N}$
$n\ge1\Rightarrow n^p\ge1\Rightarrow n^p(\sqrt{n+1}-\sqrt{n})\ge(\sqrt{n+1}-\sqrt{n})$
$\therefore\underset{n=1}{\overset{\infty}{\sum}}n^p(\sqrt{n+1}-\sqrt{n})\ge\overset{\infty}{\underset{n=1}{\sum}}(\sqrt{n+1}-\sqrt{n})\ge0$
We know that $\overset{\infty}{\underset{n=1}{\sum}}(\sqrt{n+1}-\sqrt{n})$ diverges
$\Rightarrow\underset{n=1}{\overset{\infty}{\sum}}n^p(\sqrt{n+1}-\sqrt{n})$ also diverges.
I don't know how to proceed when $p<0$.
Use the following fact: $$\sum_{n=0}^{\infty}\frac{1}{n^p}<\infty \text{ if } p> 1 $$ For $p<0$, $$\sum_{n=1}^{\infty}\frac{1}{2n^{1/2}n^{-p}}<\infty, \text{ if } \frac{1}{2}-p>1$$ and $$\sum_{n=1}^{\infty}\frac{1}{2n^{1/2}n^{-p}}=\infty, \text{ if } \frac{1}{2}-p\leq 1$$