$$\sum_{n=1}^{\infty}\frac{\sqrt{n+2}-\sqrt{n+1}}{n}$$
Where $$a_n=\frac{\sqrt{n+2}-\sqrt{n+1}}{n}$$
I want to check whether this sum converges... At quick glance this series seems to converge because $$n>\sqrt{n+2}-\sqrt{n+1}$$
The ratio test, root test, integral test did not work in this case, so I tried the limit comparison test with $$b_n=\frac{1}{n^2}$$ $$b_n=\frac{1}{n^3}$$
Basically any p-series with $p=2,3,4,5...$ i.e. an integer, gives $$\lim_{n\to\infty}\frac{a_n}{b_n}=\infty$$
Then I tried using non-integer values of $p$, e.g.: $$b_n=\frac{1}{n^{3/2}}$$
I got $$\lim_{n\to\infty}\frac{a_n}{b_n}$$ $$=\lim_{n\to\infty}\frac{n^{3/2}(\sqrt{n+2}-\sqrt{n+1})}{n}$$ $$=\lim_{n\to\infty}\sqrt{n^2+2n}-\sqrt{n^2+n}$$ $$=\lim_{n\to\infty}(\sqrt{n^2+2n}-\sqrt{n^2+n})\frac{\sqrt{n^2+2n}+\sqrt{n^2+n}}{\sqrt{n^2+2n}+\sqrt{n^2+n}}$$ $$=\lim_{n\to\infty}\frac{n}{\sqrt{n^2+2n}+\sqrt{n^2+n}}$$ $$=\lim_{n\to\infty}\frac{1}{\sqrt{1+\frac{2}{n}}+\sqrt{1+\frac{1}{n}}}=\frac{1}{2}$$
Therefore by the limit comparison test, $a_n$ converges because $b_n$ converges by the p-series test. Then I decided to try with $$b_n=\frac{1}{n^{5/2}}$$ I got $$\lim_{n\to\infty}\frac{a_n}{b_n}=\infty$$
So now I'm paranoid because I have no idea why the test works with $p=\frac{3}{2}$ and nothing else! I'm not going to have time to check different values of $p$ during an examination... My question is this: Is there an easy way to figure out which $p$ value I should choose when working with a similar series? (i.e. a series with radicals)
$$\sum_{n=1}^{\infty}\frac{\sqrt{n+2}-\sqrt{n+1}}{n} =\\ \sum_{n=1}^{\infty}\frac{\sqrt{n+2}-\sqrt{n+1}}{n}\frac{\sqrt{n+2}+\sqrt{n+1}}{\sqrt{n+2}+\sqrt{n+1}} = \\ \sum_{n=1}^{\infty}\frac{n+2-n-1}{n(\sqrt{n+2}+\sqrt{n+1})} =\\ \sum_{n=1}^{\infty}\frac{1}{n(\sqrt{n+2}+\sqrt{n+1})}.$$
The last one is $\sim \frac{1}{n^{\frac{3}{2}}}$ and thus converges.