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Math Help - Determine values for convergence

  1. #1
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    Determine values for convergence

    Hi, how do I determine the values of p for which this series converges?:

    \sum_{n=1}^{\infty} \frac{ln n}{n^p}

    Thank you.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Let's define \zeta(p) as...

    \zeta (p) = \sum_{n=1}^{\infty} \frac{1}{n^{p}}= \sum_{n=1}^{\infty} e^{-p\cdot \ln n} (1)

    ... being p a complex variable. It is well known that the series (1) converges for Re(p) >1 so that  \zeta (p) is analytic in that region of complex plain. If \zeta(p) is analytic, then its derivative...

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

    ... exists and the series in (2) also converges for Re(p)>1...

    Kind regards

    \chi \sigma
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by coldfire View Post
    Hi, how do I determine the values of p for which this series converges?:

    \sum_{n=1}^{\infty} \frac{ln n}{n^p}

    Thank you.
    It most definitely diverges for p\le1 since then \frac{\ln(x)}{x^p}\ge\frac{1}{x^p}. For p>1 consider the integral test.
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