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)]2⋅n[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)⋅[limn→∞log(n+1)log(n+2)]2=|x|[limn→∞1n+1⋅n+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 −1≤x≤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
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)]2≤∞∑n=11n[log(n+1)]2<∞.
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