Determine the value of $ p $ for which the following infinite series converges and for which it diverges. Determine the value of $ p $ for which the following infinite series converges and for which it diverges:
$$
\sum_{n = 2}^{\infty} \frac{\sqrt{n + 2} - \sqrt{n - 2}}{n^{p}}.
$$
I don’t know how to start.
 A: Hint:
$$\frac{\sqrt{n+2}-\sqrt{n-2}}{n^p} = \frac{4}{n^p(\sqrt{n+2}+\sqrt{n-2})} \sim 2n^{-(p + 1/2)}
$$
A: @RRL provided a solid approach so I thought it might be instructive to present a different way forward.  Here we use asymptotic analysis.  To that end, we write
$$\sqrt{n\pm2}=n^{1/2}\left(1\pm\frac2n\right)^{1/2}=n^{1/2}\left(1\pm\frac1n+O\left(\frac{1}{n^2}\right)\right)$$
Thus, 
$$\frac{\sqrt{n+2}-\sqrt{n-2}}{n^p}=\frac{2}{n^{p+1/2}}+O\left(\frac{1}{n^{3/2}}\right)$$
Thus, the series of interest converges for $p>1/2$ and diverges otherwise.
A: Another way can be this. We have $$\sum_{n\geq2}\frac{\sqrt{n+2}-\sqrt{n-2}}{n^{p}}=2^{1-p}+\frac{1}{2}\sum_{n\geq3}\frac{1}{n^{p}}\int_{n-2}^{n+2}\frac{1}{\sqrt{t}}dt=2^{1-p}+2\sum_{n\geq3}\frac{1}{n^{p}\sqrt{c_{n}}}
 $$ where $c_{n}\in\left[n-2,n+2\right]
 $ by the mean value theorem. Then follows that $$ 2^{1-p}+2\sum_{n\geq3}\frac{1}{n^{p}\sqrt{n+2}}\leq\sum_{n\geq2}\frac{\sqrt{n+2}-\sqrt{n-2}}{n^{p}}\leq2^{1-p}+2\sum_{n\geq3}\frac{1}{n^{p}\sqrt{n-2}}
 $$ and so $$\sum_{n\geq2}\frac{\sqrt{n+2}-\sqrt{n-2}}{n^{p}}\sim\sum_{n\geq2}\frac{1}{n^{p+1/2}}.
 $$ 
