TITLE: alternating series test of $\sum(-1)^n\frac{\sqrt{n+1}-\sqrt{n}}{n}$
QUESTION [5 upvotes]: I want to know if the sum $$\sum_{n=1}^\infty(-1)^n\frac{\sqrt{n+1}-\sqrt{n}}{n}$$ converges or not.
So I've tried the alternating series text:

$\lim\limits_{n\rightarrow\infty}\frac{\sqrt{n+1}-\sqrt{n}}{n}=0$ is clear.
I also need $a_{n+1}\leq a_n$. 

So my question is if there is an very easy way to show 2. ? (I've tried to calculate the inequation but I don't get a nice result.)

REPLY [4 votes]: The answer by @Amr is very nice and solves your problem. 
Here are some other approaches. 
I. Let $a_n =  (\sqrt{n+1}-\sqrt{n})/n$.
Then 
$$\begin{eqnarray*}
\frac{a_{n+1}}{a_n} 
&=& \frac{\sqrt{n+2}-\sqrt{n+1}}{n+1} \frac{n}{\sqrt{n+1}-\sqrt{n}} \\
&=& \frac{n}{n+1}(\sqrt{n+2}-\sqrt{n+1})(\sqrt{n}+\sqrt{n+1}) \\
&<& \frac{n}{n+1}(\sqrt{n+2}-\sqrt{n+1})(\sqrt{n+2}+\sqrt{n+1}) \\
&=& \frac{n}{n+1}. 
\end{eqnarray*}$$
II. Since 
$$\frac{a_{n+1}}{a_n} \sim 1 - \frac{3}{2n} + \ldots
\qquad (n\to\infty)$$
the series converges absolutely by Raabe's test.