Convergence or divergence of $ \sum_{n=1}^{\infty}\frac{\sqrt{n+2}-\sqrt{n}}{n^{3/2}} $: How to argue? Does the series 
$$
\sum_{n=1}^{\infty}\frac{\sqrt{n+2}-\sqrt{n}}{n^{3/2}}
$$
converge or diverge?
My attempt was to write the series as 
$$
\sum_{n=1}^{\infty}\frac{\sqrt{n+2}-\sqrt{n}}{n^{3/2}}=\sum_{n=1}^{\infty}\frac{\sqrt{n+2}}{n^{3/2}}-\sum_{n=1}^{\infty}\frac{1}{n}
$$
The first series can be estimated from below by the harmonic series:
$$
\sum_{n=1}^{\infty}\frac{\sqrt{n+2}}{n^{3/2}}=\sum_{n=1}^{\infty}\frac{(n+2)^{1/2}}{n^{1/2}\cdot n}\geqslant\sum_{n=1}^\infty \frac{1}{n}=\infty
$$
and hence diverges.
The second series is the harmonic series and hence diverges.
Now, I am not sure what the whole thing does.
 A: Your attempt is wrong because you can't separate the series into a difference of divergent series. You got an indeterminate form $\infty - \infty$.
You could try to rationalize the numerator:
$$(\sqrt{n+2}-\sqrt{n})\frac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+2}+\sqrt{n}}=\frac{2}{\sqrt{n+2}+\sqrt{n}}$$
Then the series becomes 

$$\sum_{n=1}^{\infty}\frac{2}{(\sqrt{n+2}+\sqrt{n})n^{3/2}}$$

You can use comparison test to conclude.
A: HINT
We can more effectively use that
$$\sqrt{n+2}-\sqrt{n}=\frac{2}{\sqrt{n+2}+\sqrt{n}}$$
A: You can use the estimate:
$$\frac{\sqrt{n+2}-\sqrt{n}}{n^{3/2}}<\frac1{n^2} \iff \sqrt{n}(\sqrt{n+2}-\sqrt{n})<1 \iff \\
\sqrt{n(n+2)}<n+1 \iff n^2+2n<n^2+2n+1.$$
A: Rewriting :
$2=(n+2) -n=$
$(\sqrt{n+2}-√n)(\sqrt{n+2}+√n) \gt$
$\sqrt{n+2}-√n$, since 
$\sqrt{n+2}+√n >1$.
Hence 
$\dfrac{\sqrt{n+2}-√n}{n^{3/2}} \lt \dfrac{2}{n^{3/2}}.$
Use comparison test.
A: Note that $\sqrt{n+2}-\sqrt n\to 0$ as $n\to\infty$, so $\;\dfrac{\sqrt{n+2}-\sqrt{n}}{n^{3/2}}=o\biggl(\dfrac 1{n^{3/2}}\biggr)$, and the latter series converges.
