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Thread: Convergence of complex series

  1. #1
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    Convergence of complex series

    Hi guys, I think this is easy, but I keep getting stuck!

    "Prove the series $\displaystyle \sum_{n=1}^{\infty} (1+n)^{-z}$ converges for all complex z with real part strictly greater than 1."

    I assume we take the absolute value $\displaystyle \sum_{n=1}^{\infty} |(1+n)^{-z}|$

    and then $\displaystyle \sum_{n=1}^{\infty} |(1+n)^{-Re(z)}||(1+n)^{-Im(z)}|$

    $\displaystyle \leq \sum_{n=1}^{\infty} |(1+n)^{-Re(z)}|$

    But where to now, I assume the comparison test, but what would one compare this too?

    Many thanks,

    Yaad
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  2. #2
    MHF Contributor chisigma's Avatar
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    If we consider the series...

    $\displaystyle \displaystyle S=\sum_{n=2}^{\infty} \frac{1}{n^{\text{Re} (z)}}$ (1)

    ... setting $\displaystyle \text{Re} (z) = 1+\varepsilon\ ,\ \varepsilon>0$ we have that for n 'large enough' is...

    $\displaystyle \displaystyle \frac{1}{n^{1 + \varepsilon}}} < \frac{1}{n\ \ln^{2} n}$ (2)

    Now the series...

    $\displaystyle \displaystyle \sum_{n=2}^{\infty} \frac{1}{n\ \ln^{2} n}$ (3)

    ... converges so that the series (1) for $\displaystyle \text{Re} (z)>1$ converges too...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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