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Thread: 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?:

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

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

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

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

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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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?:

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

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