real analysis - For what values of $x$ in $(-3,17)$ does the series $sumlimits^{infty}_{n=1}frac{(-1)^n x^n}{n[log (n+1)]^2}$ converge?

For what values of $x$ in the following series, does the series converge?

\begin{align}\sum^{\infty}_{n=1}\dfrac{(-1)^n x^n}{n[\log (n+1)]^2},\;\;-3

MY TRIAL

\begin{align}\lim\limits_{n\to \infty}\left|\dfrac{(-1)^{n+1} x^{n+1}}{(n+1)[\log (n+2)]^2}\cdot\dfrac{n[\log (n+1)]^2}{(-1)^n x^n}\right|&=|x|\lim\limits_{n\to \infty}\left|\dfrac{n}{n+1}\cdot\left[\dfrac{\log (n+1)}{\log (n+2)}\right]^2\right|\\&=|x|\lim\limits_{n\to \infty}\left(\dfrac{n}{n+1}\right)\cdot\lim\limits_{n\to \infty}\left[\dfrac{\log (n+1)}{\log (n+2)}\right]^2\\&=|x|\lim\limits_{n\to \infty}\left(\dfrac{n}{n+1}\right)\cdot\left[\lim\limits_{n\to \infty}\dfrac{\log (n+1)}{\log (n+2)}\right]^2\\&=|x|\left[\lim\limits_{n\to \infty}\dfrac{1}{n+1}\cdot n+2\right]^2\\&=|x|\end{align}
Hence, the series converges absolutely for $|x|<1$ and diverges when $|x|>1$.

When $x=1,$
\begin{align}\sum^{\infty}_{n=1}\dfrac{(-1)^n }{n[\log (n+1)]^2}<\infty\;\;\text{By Alternating series test}\end{align}
When $x=-1,$
\begin{align}\sum^{\infty}_{n=1}\dfrac{1}{n[\log (n+1)]^2}<\infty\;\;\text{By Direct comparison test}\end{align}

Hence, the values of $x$ for which the series converges, is $-1\leq x\leq 1.$

I'm I right? Constructive criticisms will be highly welcome! Thanks!

Answer

You are correct. This is a "variation on the theme".

For $|x|>1$
$$\lim_{n\to +\infty}\dfrac{|-x|^n}{n[\log (n+1)]^2}=+\infty$$
and the series is divergent.

For $|x|\leq 1$, by direct comparison, the series is absolutely convergent
$$\sum^{\infty}_{n=1}\dfrac{|-x|^n}{n[\log (n+1)]^2}\leq \sum^{\infty}_{n=1}\dfrac{1}{n[\log (n+1)]^2}<\infty.$$