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Math Help - find its radius of convergence

  1. #1
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    find its radius of convergence

    sigma ( n^(sqrt(n)) * z^n, n>= 1, z is complex)

    how to find its radius of convergence, it tried the ratio test
    but it doesnt work
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  2. #2
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    Quote Originally Posted by szpengchao View Post
    sigma ( n^(sqrt(n)) * z^n, n>= 1, z is complex)

    how to find its radius of convergence, it tried the ratio test
    but it doesnt work
    I think you'll find that \lim_{n \rightarrow \infty} \frac{(n+1)^{\sqrt{n+1}}}{n^{\sqrt{n}}} = 1.

    So the ratio test forces the requirement |z| < 1. Then you must test the cases z = 1 and z = -1 for convergence, which should be very simple to do .....
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    So the ratio test forces the requirement |z| < 1. Then you must test the cases z = 1 and z = -1 for convergence, which should be very simple to do .....
    No. z is complex so |z|=1\implies z = e^{i\theta}. There are infinitely many points to check.
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