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Series 4.

  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=3}^{\infty}\frac{1}{\left(\ln (\ln n)\right)^{\ln n}}
    [/tex]





    3. The attempt at a solution

    wait I'll try to solve
     
    Last edited: Nov 18, 2007
  2. jcsd
  3. Nov 18, 2007 #2
    I deleted wrong solution here.
     
    Last edited: Nov 18, 2007
  4. Nov 18, 2007 #3
    [tex]\frac{1}{\left(\ln (\ln n)\right)^{\ln n}}=\frac{1}{e^{\ln n\ln (\ln (\ln n))}}=\frac{1}{n^{\ln (\ln (\ln n))}}[/tex]

    for n enough big [tex]\ln (\ln (\ln n))>2[/tex].and we can say that series converges
     
    Last edited: Nov 18, 2007
  5. Nov 19, 2007 #4

    Gib Z

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    That seems the correct method, though in case you want to clarify it, prove that ln(ln(ln n)) > 2 and is increasing, and also show that you are comparing it to say, zeta(2).
     
  6. Nov 19, 2007 #5
    To prove that it's increasing is it enough?

    [tex]\left(\ln (\ln (\ln n))\right)'=\frac{1}{n\ln (\ln n)}>0[/tex]
     
  7. Nov 19, 2007 #6

    quasar987

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    you got the derivative wrong. think chain rule.

    and in case you're wondering zeta(s)=[itex]\sum[/itex]1/n^s, so zeta(2)=[itex]\sum[/itex]1/n². for a complete solution, you should specify that you're comparing the general term to that of zeta(2).
     
    Last edited: Nov 19, 2007
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