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Convergence of a logaritmic series

  1. Oct 24, 2013 #1
    1. The problem statement, all variables and given/known data
    Analyze the convergence of the following series, describing the criteria used:

    2. Relevant equations

    3. The attempt at a solution
    Wolfram Alpha says it converges due to comparison test, however I can't find to get a proper comparison. My main attempt was starting with [itex]\displaystyle ln(n)< n[/itex] and, starting from there, getting:
    [itex](ln(ln(n)))^{ln(n)} < (ln(n))^{ln(n)} < n^{ln(n)} < n^{n}[/itex]
    However, after getting their reciprocal, I manage to prove [itex]\frac{1}{n^{n}}[/itex] converges, but that is inconclusive. Any idea to which function I should start with to get the comparison test right?
  2. jcsd
  3. Oct 24, 2013 #2


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    If you want to show it converges, then you need to replace [itex] \ln(\ln(n))^{\ln(n)} [/itex] with something that is smaller than it, not larger. That means you need to think of a function which is smaller than ln(n) to replace some of those logs with.

    It might help to remember that whenever there's a ln(n) in the exponent, you can do some algebra on it to rewrite your function!
  4. Oct 25, 2013 #3
    All right, managed to get [itex]\displaystyle ln(ln(n))^{ln(n)} = n^{ln(ln(ln(n)))}[/itex]. Guess using [itex]n^{2}[/itex] as the comparison function would work, right? Because at infinity, [itex]n^{ln(ln(ln(n)))} ≈ n^{n}[/itex], therefore I would it should converge.
  5. Oct 25, 2013 #4


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    I don't think that it looks anything close to nn but I agree that n2 is a good comparison.
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