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Interesting Series

  1. Mar 9, 2013 #1
    1. The problem statement, all variables and given/known data

    Is this series convergent or divergent?

    [tex] \sum_{n=2}^{\infty} \frac{1}{(ln(n))^{ln \, ln(n)}} [/tex]



    2. Relevant equations

    [tex] ln(x)<x^a, \, a>0 [/tex]



    3. The attempt at a solution

    I don't see a way to tackle this one other than the comparison test.

    [tex] \sum_{n=2}^{\infty} \frac{1}{(ln(n))^{ln \, ln(n)}} < or > ? [/tex]

    What would I compare this to?
     
    Last edited: Mar 9, 2013
  2. jcsd
  3. Mar 9, 2013 #2

    Ray Vickson

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    Re: LaTeX code: you need }} instead of } to close the \frac command properly. Also, you should write '\ln' instead of 'ln'; 'ln' produces ##ln(n)##, while '\ln' gives ##\ln(n)##.
     
  4. Mar 9, 2013 #3

    Zondrina

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    Slow down and take your hint into consideration.

    For ##a>0##, if ##ln(n) < n^a## then ##(ln(n))^{ln(ln(n))} < (n^a)^{ln(ln(n))}##.

    Reciprocating both sides, you get ##\frac{1}{(ln(n))^{ln(ln(n))}} > \frac{1}{n^{aln(ln(n))}} >## something I will allow you to notice for yourself.

    Then you will have to break down the cases for positive a.... you'll need two cases in total. Pretty sure it was 'p-series' though, not 'a-series' :).
     
  5. Mar 9, 2013 #4

    I like Serena

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    Based on visual inspection it looks as if ##\frac{1}{(\ln n)^{\ln(\ln n)}} > \frac 1 {n^{0.1}} > \frac 1 n## for n sufficiently large.
    The series diverges.


    I don't see how I can match that with your observation though.
    The series ##\dfrac{1}{n^{a\ln(\ln n)}}## converges for any a>0, meaning it won't give conclusive evidence.
     
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