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Math Help - show this series defines a continuous function on (0,1]

  1. #1
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    show this series defines a continuous function on (0,1]

     \sum_{n=1}^{\infty}{\frac{1}{n^{x+1}}}
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  2. #2
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    Quote Originally Posted by silversand View Post
     \sum_{n=1}^{\infty}{\frac{1}{n^{x+1}}}
    for it not to be continuous then

     n^{x+1} = 0 on the given interval

     x+1 = \log_n(0) is undefined!

    which is not on  (0,1]

    therefore must be continuous
    Last edited by pickslides; May 3rd 2009 at 11:07 PM. Reason: typo
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  3. #3
    MHF Contributor chisigma's Avatar
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    The function can be written as...

    f(x)= \sum_{n=1}^{\infty}\frac{1}{n^{1+x}}= \sum_{n=1}^{\infty} e^{-(1+x)\cdot \ln n} (1)

    ... so that for any x_{0}>0 it exists an \epsilon with 0<|\epsilon|<x_{0} and for which ...

    \lim_{\epsilon \rightarrow 0} f(x_{0}+\epsilon)= \lim_{\epsilon \rightarrow 0} \sum_{n=1}^{\infty} e^{-(1+x_{0})\cdot \ln n}\cdot e^{-\epsilon \cdot \ln n}=

    = \lim_{\epsilon \rightarrow 0} \sum_{n=1}^{\infty} \frac{1}{n^{1+x_{0}}}\cdot \frac{1}{n^{\epsilon}}= \lim_{\epsilon \rightarrow 0} \sum_{n=1}^{\infty} \frac{1}{n^{1+x_{0}}} = f(x_{0}) (2)

    Therefore f(x) defined in (1) is a continous function...

    Kind regards

    \chi \sigma
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