find all real values of $x$ for which the series converges absolutely?

Question

The problem:

Find all real values of $x$ for which the series $$\sum_{n=2}^{\infty} \frac{x^n}{n(\log n)^2}$$ converges absolutely.

My attempt: My answer is $|x| <1$ because when $n \to \infty$, $x^ n$ tends to $0$ if $-1<x<1$

Is it correct?

Any hints/solution will be appreciated. Thank you.

I guess the series is also absolutely convergent when $x=1$ or $x=-1$ because $\sum_{n=2}^\infty \frac{1}{n(\log n)^2}$ is convergent. — John_Wick, Nov 28 '18 at 23:09

score 4 · Accepted Answer · answered Nov 28 '18 at 23:08

4

Yes that's correct indeed by root test we obtain

$$|x|\left(\frac{1}{n(\log n)^2}\right)^\frac1n\to |x|<1$$

and it converges also for $x=\pm 1$, can you see why?

answered Nov 28 '18 at 23:08

user

154,566

if $x = -1$, then it will converges by leibnitz test???? Am i right – jasmine Nov 28 '18 at 23:12
1

@jasmine Once we know that it converges for $x=1$ it converges also absolutely for $x=-1$ therefore we don't need to invoke Leibnitz. – user Nov 28 '18 at 23:13
@ gimusi that mean answer will be $[-1,1]$? Am i right ?? – jasmine Nov 28 '18 at 23:14
@jasmineYes of course. How do you prove convergence for the case $x=1$? – user Nov 28 '18 at 23:15
@ gimusii mean $\sum_{n=2}^{\infty} \frac{1}{n(logn)^2} \le \sum \frac{1}{n^2} $? we know by p- test it will converge – jasmine Nov 28 '18 at 23:17
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@jasmine Mmmhhhh...that's does not seem a good method.You are claiming that the LHS is greater than something that converges, how can we conclude from here? – user Nov 28 '18 at 23:19
@ gimusi ,,,i gots its ..thanks u – jasmine Nov 28 '18 at 23:21
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@jasmine We need integral test to prove that or Cauchy condensation test. Refer to the related OP – user Nov 28 '18 at 23:23
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@jasmine You are welcome! Be – user Nov 28 '18 at 23:24
$f(1)=2.10974280123689197448$ and $f(-1)=0.84776283970567945082$ – robjohn Nov 29 '18 at 03:48
@robjohn Thank! Are those numerical estimation? – user Nov 29 '18 at 07:34
@JyrkiLahtonen Thanks! After you have pointed out my mistakes, I was searching for some related OP and I've found that. That's really a too advanced problem for my current knowledge :( – user Nov 29 '18 at 07:37
@gimusi: yes, they are not exact, but they are accurate to the precision given. – robjohn Nov 29 '18 at 08:11
@robjohn Ah ok, I thought I was missing something else. Thank again! Bye – user Nov 29 '18 at 08:21

find all real values of $x$ for which the series converges absolutely?

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