nature of the series $\sum \tfrac{(-1)^{n}\ln(n)}{\sqrt{n+2}}$ 
I would like to prove the following series convergent 

$$\dfrac{(-1)^{n}\ln(n)}{\sqrt{n+2}},\quad \dfrac{(-1)^{n}\ln(2)}{\sqrt{n+2}}$$
using Alternating series test:


*

*$u_n=\dfrac{(-1)^{n}\ln(n)}{\sqrt{n+2}}$, so 
$|u_n|=\dfrac{\ln(n)}{\sqrt{n+2}}$. We have :  $$\dfrac{\ln(n)}{\sqrt{n+2}}\sim \dfrac{\ln(n)}{\sqrt{n}}.$$
As $\ln(n)=\mathcal{o}\left( \sqrt{n}\right)$, then $\dfrac{\ln(n)}{\sqrt{n+2}}  \underset{ \overset { n \rightarrow +\infty } {} } {\longrightarrow }0 $


Still I need to  show that $|u_n|$ is decreases monotonically:
I'm stuck here, I tired $|u_{n+1}|-|u_n|$ and $\dfrac{|u_{n+1}|}{|u_n|}$ and 
I tried to show that if $|u_n|=f(n)=\dfrac{\ln(n)}{\sqrt{n+2}}$
 then
$$f'(x)=\dfrac{2x+4-x\ln(x)}{2x\left(x+2 \right)^{\frac{3}{2}}},$$
but I can't tell if its negative or positive.


*

*$u_n=\dfrac{(-1)^{n}\ln(2)}{\sqrt{n+2}}$, so 
$|u_n|=\dfrac{\ln(2)}{\sqrt{n+2}}$ and 
$|u_n|=\dfrac{\ln(n)}{\sqrt{n+2}} \underset{ \overset { n \rightarrow +\infty } {} } {\longrightarrow }0 $

*If $|u_n|=g(n)=\dfrac{\ln(2)}{\sqrt{n+2}}$
then
$g'(n)=\frac{-1}{2\sqrt{(n+2)^{3}}}\leq 0$ then by Alternating series test 
$$\sum_{n\geq 0}\dfrac{(-1)^{n}\ln(2)}{\sqrt{n+2}}$$ is convergent.

You have a typo/mistake in $f'$: it should be 
$$
f'(x)=\dfrac{2x+4-x\ln(x)}{2x\left(x+2 \right)^{\frac{3}{2}}}.
$$
Because you only care about the sign, and the denominator is clearly positive for $x>0$, we should only look at the sign of 
$$
{2x+4-x\ln(x)}=4+x(2-\ln x).
$$
Since the expression in brackets tends to $-\infty$ as $x\to\infty$, we have that 
$$
\lim_{x\to\infty}2x+4-x\ln(x)=-\infty,
$$
and so for $x$ big enough, the function $f$ is decreasing. 
