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Find Radius of Convergence of this Complex Series

  1. Apr 24, 2010 #1
    Hi, am a bit stuck with this.

    Find the radius of convergence of the complex series

    (Sigma n=1 to infinity) (z - e)^n!

    I know that the answer is R=1 but I'm not sure how to get there.
    It's the factorial as a power which I'm not sure about, have seen this in some other problems too.
    I have tried using the Ratio Test to determine radius of convergence but doesn't seem to be working.

    Any ideas anyone?

    Thanks in advance, and apologies for the lack of Latex.
  2. jcsd
  3. Apr 24, 2010 #2


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    The ratio test should work. The ratio of the n term to the (n-1) term is (z-e)n.
    If |z-e| < 1, the ratio -> 0, while if |z-e| > 1, tie ratio becomes infinite.
  4. Apr 25, 2010 #3
    Thanks, though I'm still not 100% sure about how to get this.

    Using ratio test I obtain

    mod[ (z-e)^(n+1)! / (z-e)^n! ]
    = mod[ (z-e)^n!(n+1-1) ]
    = mod[ (z-e)^n.n! ]

    Then I'm not sure where to go from here.
  5. Apr 25, 2010 #4


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    mod[ (z-e)^n.n! ] = [mod(z-e)]^n.n!

    When mod(z-e) < 1, the large exponent makes the expression very small, while if mod(x-e) > 1, it makes the expression very large. Therefore the radius of convergence = 1.
    Last edited: Apr 25, 2010
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