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



limn|(1)n+1xn+1(n+1)[log(n+2)]2n[log(n+1)]2(1)nxn|=|x|limn|nn+1[log(n+1)log(n+2)]2|=|x|limn(nn+1)limn[log(n+1)log(n+2)]2=|x|limn(nn+1)[limnlog(n+1)log(n+2)]2=|x|[limn1n+1n+2]2=|x|


Hence, the series converges absolutely for |x|<1 and diverges when |x|>1.



When x=1,
n=1(1)nn[log(n+1)]2<By Alternating series test


When x=1,
n=11n[log(n+1)]2<By Direct comparison test


Hence, the values of x for which the series converges, is 1x1.



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
limn+|x|nn[log(n+1)]2=+


and the series is divergent.




For |x|1, by direct comparison, the series is absolutely convergent
n=1|x|nn[log(n+1)]2n=11n[log(n+1)]2<.


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