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Math Help - radius of complex convergence

  1. #1
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    radius of complex convergence

    I stumbled at the answer of a problem solved in Reily-Hobson-Bence.Please check if I am wrong:

    Find the parts of the z-plane for which the following series is convergent:

    ∑[(1/n!)(z^n)] where n runs from 0 to ∞


    Cauchy's radius: (1/R)=Lt (n->∞) [(1/n!)^(1/n)]

    Changing variables as 1/n=t,the limit becomes (1/R)=Lt(t->0) [(t!)^t]
    I am having the limit as 1 and hence,R=1...but the book says Since
    [(n!)^(1/n)] behaves like n as n → ∞ we find lim[(1/n!)^(1/n)] = 0. Hence R = ∞ and the series is convergent for all z
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  2. #2
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    Quote Originally Posted by kolahalb View Post
    I stumbled at the answer of a problem solved in Reily-Hobson-Bence.Please check if I am wrong:

    Find the parts of the z-plane for which the following series is convergent:

    ∑[(1/n!)(z^n)] where n runs from 0 to ∞


    Cauchy's radius: (1/R)=Lt (n->∞) [(1/n!)^(1/n)]

    Changing variables as 1/n=t,the limit becomes (1/R)=Lt(t->0) [(t!)^t]
    I am having the limit as 1 and hence,R=1...but the book says Since
    [(n!)^(1/n)] behaves like n as n → ∞ we find lim[(1/n!)^(1/n)] = 0. Hence R = ∞ and the series is convergent for all z
    \frac{1}{n!} \ne (1/n)! whatever you think (1/n)! might mean for n \in \mathbb{N}.

    RonL
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  3. #3
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    Yes,I see that. But how to see that [(n!)^(1/n)] behaves like n as n → ∞
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by kolahalb View Post
    Yes,I see that. But how to see that [(n!)^(1/n)] behaves like n as n → ∞
    Replace n! by Stirlings approximation and it will appear.

    <br />
n! \sim \sqrt{2 \pi n} n^n e^{-n}<br />

    RonL
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  5. #5
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    Thank you for your help.

    I resolved the problem in this manner: (1/n!) falls at a much faster rate than that of (1/n).As a result the function is decreasing all the way starting from the highest value one and always being positive.If you draw a graph,the curve should tend to zero as n tends to infinity...
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  6. #6
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    You can use the root/ratio test.
    Define x_n = n!.
    Since \lim \frac{x_{n+1}}{x_n} = +\infty it means \lim |x_n|^{1/n} = \lim (n!)^{1/n} = \infty.
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  7. #7
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    Yes,I know that ratio test helps better;but I was having trouble with the question I posted.In the book the problem was solved using Cauchy's root test.
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