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Math Help - Radius of convergence

  1. #1
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    Post Radius of convergence

    How do I find R when the power series has n! as the power of x?

    Thanks for helping out.
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  2. #2
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    Quote Originally Posted by DaRush19 View Post
    How do I find R when the power series has n! as the power of x?
    If you want help, you give the exact question.
    There is no point in our guessing.
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  3. #3
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    e.g Find the radius of convergence of the sum of 1/n! x^n!
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  4. #4
    MHF Contributor chisigma's Avatar
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    For |x| \le 1 is |x|^{n!} \le |x|^{n} so that...

    \sum_{n=0}^{\infty} \frac{|x|^{n!}}{n!} \le \sum_{n=1}^{\infty} \frac{|x|^{n}}{n!} = e^{|x|} (1)

    ... and the series of powers of x converges. If |x|>1 we set |x|= 1 + \delta, \delta >0 so that is...

    \frac{|x|^{n!}}{n!} = \prod_{k=1}^{n} \frac{(1+\delta)^{k}}{k}\rightarrow \ln (\frac{|x|^{n!}}{n!} ) = \ln (1+\delta) \sum _{k=1}^{n} k - \sum_{k=1}^{n} \ln k (2)

    But is  \ln (1+\delta) >0 so that is...

    \lim_{n \rightarrow \infty} \frac{|x|^{n!}}{n!} = + \infty (3)

    ... and the series of powers diverges. Then it seems to be R=1...

    Kind regards

    \chi \sigma
    Last edited by chisigma; September 7th 2009 at 02:08 AM. Reason: minor editing details...
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