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Math Help - Simple Convergence limit

  1. #1
    Junior Member RaisinBread's Avatar
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    Simple Convergence limit

    Hey, I'm trying to show that the following sequence of functions
    f_n(x)=\sum_{k=1}^{n}\frac{1}{(x+k)^2}
    converges uniformly on \mathbb{R}^+.

    To do that I'm trying to find the limit of the simple convergence of the sequence, I can easily show that the limit for the simple convergence exists by noticing that for every x\in\mathbb{R}^+,
    0\leq\sum_{n=1}^{\infty}\frac{1}{(x+n)^2}\leq\sum_  {n=1}^{\infty}\frac{1}{n^2}
    and use the convergence of the Riemann Zeta Function. However I can't manage to what the function of the limit is.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Simple Convergence limit

    Quote Originally Posted by RaisinBread View Post
    Hey, I'm trying to show that the following sequence of functions
    f_n(x)=\sum_{k=1}^{n}\frac{1}{(x+k)^2}
    converges uniformly on \mathbb{R}^+.

    To do that I'm trying to find the limit of the simple convergence of the sequence, I can easily show that the limit for the simple convergence exists by noticing that for every x\in\mathbb{R}^+,
    0\leq\sum_{n=1}^{\infty}\frac{1}{(x+n)^2}\leq\sum_  {n=1}^{\infty}\frac{1}{n^2}
    and use the convergence of the Riemann Zeta Function. However I can't manage to what the function of the limit is.
    In...

    http://www.mathhelpforum.com/math-he...-i-188482.html

    ... it has been demonstrated that...

    \sum_{n=1}^{\infty} \frac{1}{(n+a)\ (n+b)}= \frac{\phi(b)-\phi(a)}{b-a} (1)

    ... where...

    \phi(x)= \frac{d}{dx} \ln x! (2)

    ... being x! the so called 'factorial function'...

    x!= \int_{0}^{\infty} t^{x}\ e^{-t}\ dt (3)

    Now from (1) using l'Hopital rule You obtain...

    \sum_{n=1}^{\infty} \frac{1}{(x+n)^{2}} = \lim_{\xi \rightarrow x} \sum_{n=1}^{\infty} \frac{1}{(\xi+n)\ (x+n)} =\lim_{\xi \rightarrow x} \frac{\phi(\xi)-\phi(x)}{\xi-x} = \phi^{'} (x) (4)

    Kind regards

    \chi \sigma
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