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Math Help - Finding a convergence radius

  1. #1
    Super Member Showcase_22's Avatar
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    Finding a convergence radius

    Find the convergence radius of \sum_{n=0}^{\infty} n! \pi^{\frac{n}{2}}x^{n!}

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    My attempt:

    Define a_n:= n! \pi^{\frac{n}{2}}x^{n!}

    Therefore \left| \frac{a_{n+1}}{a_n} \right|=\left| \frac{(n+1)!\pi^{\frac{n+1}{2}}x^{(n+1)!}}{n! \pi^{\frac{n}{2}} x^{n!}} \right|

    =\left| (n+1) \pi^{\frac{1}{2}} x^{n! n} \right|

    If |x|<1 then R=\infty.
    If |x|\geq 1 then R=0

    Is this right?
    Last edited by Showcase_22; May 28th 2009 at 08:17 AM.
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  2. #2
    Member
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    Well I think you should have that |\frac{a_{n+1}}{a_n}|=c_n=|(n+1)\sqrt{\pi}x^{n+1}|.

    So you need \lim c_n<1\Longrightarrow \lim_n|nx^n|<\frac{1}{\sqrt{\pi}}. Which is only true if |x|<1. Thus the radius of convergence is R=1.
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