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Math Help - a Series of functions : Does it converge? Does it uniformly converge?

  1. #1
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    a Series of functions : Does it converge? Does it uniformly converge?

    Hello, I have a question:

    \sum_{n=1}^{\infty} \ln (1+\frac{x^2}{n\ln^2(n)}), x \in (-a,a)

    Does it converge? If so, does it uniformly converge?

    Thought of turning the ln into two different ln's, but I really can't find a way to prove a convergence of anything here...

    Can you please help me?

    Thanks
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    I have...

    lny<yfor  y>3<br />
    ln(1+\frac{x^2}{nln^2(n)})<1+\frac{x^2}{nln^2(n)}

    The last term is monotonic decreasing so we can use the integral test... hmmm...


    still thinking about this...
    Last edited by Also sprach Zarathustra; June 25th 2010 at 08:49 AM.
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  3. #3
    MHF Contributor chisigma's Avatar
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    The term \frac{x^{2}}{n\ \ln^{2} n} is a nonsense for n=1 , so that the series probably is...

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

    The convergence of the 'infinite sum' (1) is equivalent to the convergence of the infinite product...

    \displaystyle e^{\Lambda} = \prod _{n=2}^{\infty} (1+\frac{x^{2}}{n\ \ln^{2} n}) (2)

    ... and that is guaranted by the convergence of the series...

    \displaystyle \sum_{n=2}^{\infty} \frac{x^{2}}{n\ \ln^{2} n} (3)

    ... which is true for any value of x. The series (3) converges uniformely in any finite interval of the x and the same holds for the series (1)...

    Kind regards

    \chi \sigma
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  4. #4
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    Beautiful, thank you very much !

    Only one little thing - how do you explain the "... and that is guaranted by the convergence of the series..." (from (2) to (3) )?
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  5. #5
    MHF Contributor chisigma's Avatar
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    The convergence of the series...

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

    ... can be demostrated applying the 'integral test' considering that is...

    \displaystyle \sum_{n=2}^{\infty} \frac{1}{n\ \ln^{2} n} < 2\ \int_{2}^{\infty} \frac{dx}{x\ \ln^{2} x} = 2\ |- \frac{1}{\ln x}|_{2}^{\infty} = \frac{2}{\ln 2} (2)

    Kind regards

    \chi \sigma
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