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Math Help - Convergence of power series

  1. #1
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    Convergence of power series

    basic complex analysis (so z is complex):

    Prove that the power series $\sum_{k=0}^{\infty} z^{k}$ converges at no point on its circle of convergence |z| = 1

    Attempt:

    I have no idea what to do. This course is driving me nuts. I know for |z| < 1 that series converges to 1/1-z but that is all I know. Can anyone give me a hint?
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  2. #2
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    It is well known that geometric series are divergent where \displaystyle |r| \geq 1.

    If you must prove it though, since \displaystyle |z| = 1, then we can write \displaystyle z = e^{i\theta}.


    So \displaystyle \sum_{k = 0}^{\infty}z^k = \sum_{k = 0}^{\infty}\left(e^{i\theta}\right)^k

    \displaystyle = \sum_{k = 0}^{\infty}e^{ik\theta}

    \displaystyle = \lim_{n \to \infty}\left[\frac{e^{i\theta}\left(e^{i(n+1)\theta} - 1\right)}{e^{i\theta} - 1}\right].

    Can this limit be evaluted? If so, what does it go to?
    Last edited by Prove It; March 28th 2011 at 05:51 PM.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mulaosmanovicben View Post
    basic complex analysis (so z is complex):

    Prove that the power series $\sum_{k=0}^{\infty} z^{k}$ converges at no point on its circle of convergence |z| = 1

    Attempt:

    I have no idea what to do. This course is driving me nuts. I know for |z| < 1 that series converges to 1/1-z but that is all I know. Can anyone give me a hint?
    You know that if |z|=1 that \displaystyle z=e^{i\theta} so that you're question reduces to \displaystyle \sum_{k=0}^{\infty}e^{ik\theta} but a quick check shows that (assuming z\ne 1 but that case is trivial) \displaystyle \sum_{k=0}^{m}e^{ik\theta}=\frac{1-e^{i(m+1)\theta}}{1-e^{i\theta}} and so to assume that \displaystyle \sum_{k=0}^{\infty}z^k converges is to assume that \displaystyle \lim_{m\to\infty}e^{i(m+1)\theta} exists...I think you can take it from here.
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    Quote Originally Posted by Prove It View Post
    It is well known that geometric series are divergent where \displaystyle |r| \geq 1.

    If you must prove it though, since \displaystyle |z| = 1, then we can write \displaystyle z = e^{i\theta}.


    So \displaystyle \sum_{k = 0}^{\infty}z^k = \sum_{k = 0}^{\infty}\left(e^{i\theta}\right)^k

    \displaystyle = \sum_{k = 0}^{\infty}e^{ik\theta}

    \displaystyle = \lim_{n \to \infty}\left[\frac{e^{i\theta}\left(e^{i(n+1)\theta} - 1\right)}{e^{i\theta} - 1}\right].

    Can this limit be evaluted? If so, what does it go to?

    Why is the summation equal to that limit?
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  5. #5
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    Because for a finite geometric series, the sum of the first \displaystyle n terms is \displaystyle \frac{a\left(r^n - 1\right)}{r - 1}.

    The infinite sum will be the limit of this as \displaystyle n \to \infty.
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    An alternative:

    If |z|=1 then, |z^k|=1 which implies l=\lim_{k\to +\infty}z^k does no exist or l\neq 0 . In both cases, the necessary condition for he convergence of \sum_{k\geq 0}z^k is not satisfied.
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