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Math Help - Help with condensation test

  1. #1
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    Help with condensation test

    i guys.
    i need to determine if this serie converges:

    \sum_{n=2}^{\infty}\frac{1}{n\ln(n)\ln(\ln(n))}

    this is what i've tried so far:

    i use condensation test and got the following:

    \sum \frac{2^n}{2^n\ln2^n[\ln(\ln2^n)]}=\frac{1}{n\ln2[\ln(n\ln2)]}

    \frac{1}{\ln2}\sum \frac{1}{n[\ln(n\ln2)]}=\frac{1}{n[\ln n+\ln(\ln2)]}=\frac{1}{n\ln n+n\ln(\ln2)}

    \frac{1}{\ln2}\sum \frac{1}{n\ln n+n\ln(\ln2)}=\frac{1}{n\ln n}+\frac{1}{n\ln(\ln2)}


    now, my question is this:

    can i say that since: \sum \frac{1}{n\ln(\ln2)} does not converge the the whole thing does'nt converge?

    is there more elegant way to handle this horrible serie?

    thanks in advanced!
    Last edited by Stormey; January 21st 2013 at 09:52 AM.
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  2. #2
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    Re: Help with condensation test

    Hey Stormey.

    Have you come across the Integral Test for convergence? I would suggest looking into that with the appropriate substitution.
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  3. #3
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    Re: Help with condensation test

    Yes, chiro is right, the integral test works.

    - Hollywood

    Here is a link to the condensation test: Cauchy condensation test - Wikipedia, the free encyclopedia
    Last edited by hollywood; January 21st 2013 at 07:18 PM.
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  4. #4
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    Re: Help with condensation test

    hi, and thanks for the help.

    we didn't learn the integral test yet, so i'm supposed to answer it without it, i guess...
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  5. #5
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    Re: Help with condensation test

    Maybe this condensation test will do it.

    \sum_{n=2}^{\infty}\frac{1}{n\ln(n)\ln(\ln(n))} =

    which converges if and only if the following series converges:

    \sum_{n=2}^{\infty}\frac{2^n}{2^n\ln(2^n)\ln(\ln(2  ^n))} =

    \sum_{n=2}^{\infty}\frac{1}{n\ln(2)\ln(n\ln(2))} =

    \frac{1}{ln(2)}\sum_{n=2}^{\infty}\frac{1}{n(\ln(n  )+\ln(\ln(2)))}

    This is what you did in your original post. Now let's do the condensation test again. Our original series converges if and only if the following series converges:

    \frac{1}{ln(2)}\sum_{n=2}^{\infty}\frac{2^n}{2^n(\  ln(2^n)+\ln(\ln(2)))} =

    \frac{1}{ln(2)}\sum_{n=2}^{\infty}\frac{1}{n\ln(2)  +\ln(\ln(2))}

    We think this diverges, so we simplify by making it smaller. We have \ln(ln(2))\le 2\ln(2)\le n\ln(2), so our new sum is greater than or equal to:

    \frac{1}{ln(2)}\sum_{n=2}^{\infty}\frac{1}{n\ln(2)  + n\ln(2)}=

    \frac{1}{2ln(2)}\sum_{n=2}^{\infty}\frac{1}{n\ln(2  )}=

    \frac{1}{2(ln(2))^2}\sum_{n=2}^{\infty}\frac{1}{n}

    which diverges.

    - Hollywood
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  6. #6
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    Re: Help with condensation test

    brilliant!
    thank you very much Hollywood.

    didn't know i can use this test twice like that.
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