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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<x<17 \end{align}

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!

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  • $\begingroup$ For me it's right $\endgroup$ – Atmos Sep 8 '18 at 7:36
  • $\begingroup$ Fine for me too. I just wouldn't speak of alternating series test, because the series is in fact absolutely convergent. Why use a precision tool when you can use a hammer ? :-) $\endgroup$ – Nicolas FRANCOIS Sep 8 '18 at 7:40
  • $\begingroup$ @Nicolas FRANCOIS: Smiles... $\endgroup$ – Micheal Sep 8 '18 at 7:41
  • $\begingroup$ @Nicolas FRANCOIS:You mean the series \begin{align}\sum^{\infty}_{n=1}\dfrac{(-1)^n }{n[\log (n+1)]^2}\end{align} converges absolutely for $x=1?$ How? $\endgroup$ – Micheal Sep 8 '18 at 7:42
  • $\begingroup$ @Nicolas FRANCOIS:: Oh, I see! Robert Z has cleared my confusion! Thanks! $\endgroup$ – Micheal Sep 8 '18 at 7:46
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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 $$\sum^{\infty}_{n=1}\dfrac{|-x|^n}{n[\log (n+1)]^2}\leq \sum^{\infty}_{n=1}\dfrac{1}{n[\log (n+1)]^2}<\infty.$$

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  • $\begingroup$ Thanks, I'm grateful! $\endgroup$ – Micheal Sep 8 '18 at 7:44
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Yes it is correct, for the limit from here we can proceed as follow

$$\ldots=|x|\lim\limits_{n\to \infty}\left|\dfrac{n}{n+1}\cdot\left[\dfrac{\log (n+1)}{\log (n+2)}\right]^2\right| =|x|\lim\limits_{n\to \infty}\dfrac{1}{1+1/n}\cdot\left[\dfrac{\log n+\log (1+1/n)}{\log n+\log (1+2/n)}\right]^2=|x|\cdot1=|x|$$

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  • $\begingroup$ Thanks a lot! I appreciate! $\endgroup$ – Micheal Sep 8 '18 at 7:45

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