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Series 5.

  1. Nov 18, 2007 #1
    1. The problem statement, all variables and given/known data
    Determine whether the series converges or diverges.

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





    3. The attempt at a solution

    for [tex]u=\ln (\ln n)[/tex]

    [tex]\sum_{u=\ln (\ln 3)}^{\infty}\frac{1}{e^{u^2}}[/tex]

    from Root Test

    [tex]\lim_{u\rightarrow\infty}\sqrt{\frac{1}{e^{u^2}}}=\lim_{u\rightarrow\infty}\frac{1}{e^u}=0<1[/tex]
    series converges
     
  2. jcsd
  3. Nov 18, 2007 #2

    quasar987

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    That u substitution does not give back the original series. ln(ln(3))+1 does noe equal ln(ln(4)).

    Try simply the comparison criterion. Hint: (logx)²/x -->0, so (logx)²<x for x large enough.
     
  4. Nov 18, 2007 #3
    [tex]\ln^2x<x[/tex]

    for x=lnn

    [tex]\ln^2(\ln n)<\ln n[/tex]

    [tex]e^{\ln^2(\ln n)}<e^{\ln n}=n[/tex]

    [tex]e^{\ln^2(\ln n)}<n\rightarrow \frac{1}{e^{\ln^2(\ln n)}}>\frac{1}{n}[/tex]

    diverges
     
  5. Nov 18, 2007 #4

    quasar987

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