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Determining the convergence/divergence of this series

  1. Mar 19, 2013 #1
    I am required to determine the convergence/divergence of the following series:
    $$ \sum_{n = 2}^{\infty} \frac{1}{(ln(n))^{ln(n)}}$$

    Which test should I use? Wolfram Alpha says that the comparison test was used to determine that it was convergent, but I have no idea what series I should compare it to.

    It also said that the root/ratio tests gave inconclusive results.

    Thanks!
     
    Last edited: Mar 19, 2013
  2. jcsd
  3. Mar 19, 2013 #2

    Mark44

    Staff: Mentor

    How does n play a role here?
     
  4. Mar 19, 2013 #3
    Apologies, it should be as x approaches infinity...

    Thanks for the fix!
     
  5. Mar 19, 2013 #4

    Mark44

    Staff: Mentor

    Then this is what you want.
    $$ \sum_{n = 2}^{\infty} \frac{1}{(ln(n))^{ln(n)}}$$
     
  6. Mar 19, 2013 #5
    That's correct
     
  7. Mar 19, 2013 #6
    Integral test! (joking)

    First, rewrite the sum using ##e^{\ln z} = z##. Then use the fact that after some finite value of ##n##, ##\ln n > a, \; a>0##. Here, I would just pick a value of ##a##, let's use 2. Then:

    [tex] \sum_{n=2}^{\infty} \frac{1}{{\ln n}^{\ln n} } = \sum_{n=0}^{\infty} \frac{1}{e^{\ln({\ln n}^{\ln n}) } } = \sum_{n=2}^{\infty} \frac{1}{e^{{\ln n}^{\ln \ln n}} } = \sum_{n=2}^\infty \frac{1}{n^{\ln \ln n}} [/tex]

    By some algebraic manipulations, ## \ln n > 2, \, \forall n>e^2 \Longrightarrow \ln \ln n > \ln 2 \Longrightarrow n^{\ln \ln n} > n ^ {\ln 2} \Longrightarrow \frac{1}{n^{\ln \ln n}} < \frac{1}{n^{\ln 2}} ##

    Can you finish it from here?
     
  8. Mar 19, 2013 #7
    Thanks for your help. If I'm not mistaken, isn't ## \frac{1}{n^{\ln 2}} ## divergent? Or is there another series that I should be comparing to?
     
  9. Mar 19, 2013 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, it is divergent. piercebeatz was maybe rushing it a little. Try to use the sort of things piercebeatz was using and try to work it out for yourself. Try comparing with a p series 1/n^p. And remember the comparison doesn't have to work for all n. Just for sufficiently large n. p=ln(2) is too small for convergence. There are other choices.
     
    Last edited: Mar 19, 2013
  10. Mar 19, 2013 #9
    Sorry, didn't realize that 2 was a bad choice for a. just let a=e^2 or something like that.
     
  11. Mar 19, 2013 #10

    Dick

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    Science Advisor
    Homework Helper

    I would give TogoPogo some time to work this out solo. But sure, your solution was fine in general.
     
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