Sunday, 2 March 2014

real analysis - For what values of x in (3,17) does the series sumlimitsinftyn=1frac(1)nxnn[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



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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