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Math Help - ln convergence question

  1. #1
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    ln convergence question

    \sum_{1}^{\infty}\frac{ln k}{k\sqrt {k}}

    how to shpw if it converges or diverges
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  2. #2
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    Re: ln convergence question

    Try the integral test. The integral is convergent.
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  3. #3
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    Re: ln convergence question

    but i have trouble with the integral
    lnk=t
    1/k=dt

    but in the denominator another sqrt(k)
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  4. #4
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    Re: ln convergence question

    You need to use integration by parts with \displaystyle u = x^{-\frac{1}{2}} and \displaystyle dv = \frac{\ln{x}}{x}\,dx
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    Re: ln convergence question

    Quote Originally Posted by transgalactic View Post
    \sum_{1}^{\infty}\frac{ln k}{k\sqrt {k}} converges or diverges
    This can be done by direct comparison.

    Can you show that \ln (k) \leqslant 4\sqrt[4]{k}~?
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  6. #6
    Grand Panjandrum
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    Re: ln convergence question

    Quote Originally Posted by transgalactic View Post
    \sum_{1}^{\infty}\frac{ln k}{k\sqrt {k}}

    how to shpw if it converges or diverges
    \log(k) goes to infinity more slowly than any power of k so for k large enough there exists a positive constant A such that \log(k)/k^{1.5}<A/k^{1.4} ...
    CB
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  7. #7
    MHF Contributor chisigma's Avatar
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    Re: ln convergence question

    For \text{Re}\ s>1 the 'Riemann zeta function' is defined as...

    \zeta(s)= \sum_{k=1}^{\infty} \frac{1}{k^{s}} (1)

    ... and its derivative is...

    \zeta^{'}(s)= - \sum_{k=1}^{\infty} \frac{\ln k}{k^{s}} (2)

    ... so that is...

    \sum_{k=1}^{\infty} \frac{\ln k}{k^{\frac{3}{2}}}= - \zeta^{'}(\frac{3}{2}) (3)

    Kind regards

    \chi \sigma
    Last edited by chisigma; July 12th 2011 at 12:11 PM. Reason: error of sign... very sorry!...
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