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Math Help - Infinite Series (13).

  1. #1
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    Infinite Series (13).

    Determine the Convergence/Divergence of the following series:
    \sum_{n=2}^{\infty} \frac{1}{(ln(n))^{ln(n)}}

    Sure, It will be solved by the comparison tests.

    But its hard to find a series which works in in these tests.

    And sorry for my bad english
    Last edited by TWiX; January 31st 2010 at 04:56 PM. Reason: it is start from n=2 not 1 :D
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by TWiX View Post
    Determine the Convergence/Divergence of the following series:
    \sum_{n=2}^{\infty} \frac{1}{(ln(n))^{ln(n)}}

    Sure, It will be solved by the comparison tests.

    But its hard to find a series which works in in these tests.

    And sorry for my bad english
    I know you will hate this...but Cauchy's condensation test. \frac{2^n}{\ln^{\ln(2^n)}\left(2^n\right)}=..

    Alternatively, \frac{1}{\ln^{\ln(n)}(n)}=\text{exp}\left(\ln\left  (\frac{1}{\ln^{\ln(n)}(n)}\right)\right)=\text{exp  }\left(-\ln(n)\ln(\ln(n))\right) =\text{exp}\left(\ln\left(\frac{1}{n^{\ln(\ln(n))}  }\right)\right)=\frac{1}{n^{\ln(\ln(n))}}, and clearly n^{\ln(\ln(n))}\geqslant n^2\text{, }n\geqslant e^{e^2}.
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    I know you will hate this...but Cauchy's condensation test. \frac{2^n}{\ln^{\ln(2^n)}\left(2^n\right)}=..

    Alternatively, \frac{1}{\ln^{\ln(n)}(n)}=\text{exp}\left(\ln\left  (\frac{1}{\ln^{\ln(n)}(n)}\right)\right)=\text{exp  }\left(-\ln(n)\ln(\ln(n))\right) =\text{exp}\left(\ln\left(\frac{1}{n^{\ln(\ln(n))}  }\right)\right)=\frac{1}{n^{\ln(\ln(n))}}, and clearly n^{\ln(\ln(n))}\geqslant n^2\text{, }n\geqslant e^{e^2}.
    Yes i hate that cauchy
    nerd man .. replace every n by 2^n and check for the new one
    OMG
    it makes it two problems not only one
    NERD CAUCHY!

    Thanks for the second solution.
    I got it
    am math professor
    and omg, you did not get a headache when you writing this in latex ?
    Many exponents
    Thank you ^_^
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