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Math Help - [SOLVED] Question about radius of convergence of a series

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Question about radius of convergence of a series

    Consider the series \sum _{n=0}^{\infty} (-1)^n z^{2n+1}=z-z^3+z^5-z^7+...
    We notice that a_k=0 if k=2n and a_k=(-1)^n if k=2n+1, thus \nexists \lim |a_n|^{\frac{1}{n}}=L.
    However the series has a radius of convergence.
    But if I'm not wrong, the radius of convergence is defined as R=\frac{1}{\lim |a_n|^{\frac{1}{n}}}. So how can R exist when L does not exist?
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  2. #2
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    If z=1 then \sum_{k=0} ^{ \infty} (-1)^k z^{2k+1} does not converge, but \vert  \sum_{k=0} ^{ \infty} (-1)^k z^{2k+1} \vert \leq  \sum_{k=0} ^{ \infty} \vert (-1)^k z^{2k+1} \vert \leq  \sum_{k=0} ^{ \infty} \vert z^{k} \vert and this last one converges whenever \vert z \vert <1 so the radius of convergence is 1.
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  3. #3
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Jose27 View Post
    If z=1 then \sum_{k=0} ^{ \infty} (-1)^k z^{2k+1} does not converge, but \vert  \sum_{k=0} ^{ \infty} (-1)^k z^{2k+1} \vert \leq  \sum_{k=0} ^{ \infty} \vert (-1)^k z^{2k+1} \vert \leq  \sum_{k=0} ^{ \infty} \vert z^{k} \vert and this last one converges whenever \vert z \vert <1 so the radius of convergence is 1.
    I get what you mean, all is fine.
    But still, I don't understand why R is defined as \frac{1}{\lim |a_n|^{\frac{1}{n}}}, it doesn't make sense. Otherwise R wouldn't exist if L doesn't exist, but this is not the case.

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  4. #4
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    The definition of the radius of convergence is: If s= \limsup {\vert a_n \vert ^{ \frac{1}{n} }} then R= \frac{1}{s} and notice that in the series you got s=1.
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