How to see if $\sum_{n=1}^\infty\frac{\sqrt{n+1}-\sqrt{n}}{n}$ converges I want to see if $$\sum_{n=1}^\infty\frac{\sqrt{n+1}-\sqrt{n}}{n}$$
converges or not.

I start with $$α_{n}=\frac{\sqrt{n+1}-\sqrt{n}}{n}=$$
$$\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{n(\sqrt{n+1}+\sqrt{n})}=$$
$$\frac{n+1-n}{n(\sqrt{n+1}+\sqrt{n})}=$$
$$\frac{1}{n(\sqrt{n+1}+\sqrt{n})}.$$

And here I am stuck. How can I tell if $α_{n}$ converges or not?

In my textbook, it says that the last equation is $\leq\frac{1}{2n^{\frac{3}{2}}} $

But I cannot understand how or why it gets to that result.
We have 

$$\frac{1}{n(\sqrt{n+1}+\sqrt{n})}\le \frac{1}{n(\sqrt{n}+\sqrt{n})}=\frac1{2n^{3/2}}$$
so apparently there is a typo in your textbook.
Because $\sqrt{n+1}&gt;\sqrt n$ in the range of $n$, $\sqrt{n+1}+\sqrt n &gt; 2\sqrt n$
and therefore the ratio is $\le \frac{1}{2n^{3/2}} = \zeta(3/2)/2$, where the
Riemann zeta function has a pole at 1 but is converging for larger values.