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A tough convergence problem

  1. Oct 29, 2008 #1
    1. The problem statement, all variables and given/known data
    Determine whether the series is convergent or divergent.

    Sum from n=2 to n = infinity of

    1/[(lnx)^lnx]

    2. Relevant equations



    3. The attempt at a solution

    So far I've tried the comparison test, but all I can reach is that the series is less than the harmonic series, which of course doesn't help. Any help would be appreciated greatly, thanks.
     
  2. jcsd
  3. Oct 29, 2008 #2
    did you mean here

    [tex]\sum_{n=2}^{\infty}\frac{1}{[ln(n)]^{ln(n)}}[/tex]
     
  4. Oct 29, 2008 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Let's try comparing it with 1/n^2. Remember, you can discard any finite number of terms when you do a comparison test. Try to figure out the limit n->infinity n^2/(log(n)^log(n)). Hint: take the log of the ratio.
     
  5. Oct 30, 2008 #4
    There's a theorem said,
    if a1>=a2>=.....>=0, then [tex]\sum^{\infty}_{n=1}a_{n}[/tex] converges if and only if [tex]\sum^{\infty}_{k=0} 2^{k}a_{2^{k}}[/tex] converges.
    This will make the original series to a better-looking one
     
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