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
lim
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.
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