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.
 A: Note that the first derivative is actually:
$$\frac{2x + 4 - x \ln x}{x (x+2)^{3/2}}$$
Consider $g(x) = (2 - \ln x)x + 4$; $g'(x) = 1 - \ln x < 0$ for $x \ge e$, and (say) $g(e^3) = 4-e^3 < 0$, so for $n \ge \lceil e^3 \rceil = 21$, $u_n$ is decreasing and we can apply the AST
A: 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. 
A: Hint:
There's an error in the derivative. It's 
$$f'(x)=\frac{2x+4-x\ln x}{2x(x+2)^{3/2}}.$$
For the sign of $f'(x)$: if $x>4$, $\;2x+4\le 3x$, so $f'(x)$, which has the sign of $2x+4-x\ln x$, which is $\le x(3-\ln x)$, is negative if $x> \mathrm e^3,\;$ a fortiori if $\;x\ge 21$.
