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Thread: 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 $\displaystyle $\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 $\displaystyle \displaystyle |r| \geq 1$.

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


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

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

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

    Can this limit be evaluted? If so, what does it go to?
    Last edited by Prove It; Mar 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 $\displaystyle $\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 $\displaystyle |z|=1$ that $\displaystyle \displaystyle z=e^{i\theta}$ so that you're question reduces to $\displaystyle \displaystyle \sum_{k=0}^{\infty}e^{ik\theta}$ but a quick check shows that (assuming $\displaystyle z\ne 1$ but that case is trivial) $\displaystyle \displaystyle \sum_{k=0}^{m}e^{ik\theta}=\frac{1-e^{i(m+1)\theta}}{1-e^{i\theta}}$ and so to assume that $\displaystyle \displaystyle \sum_{k=0}^{\infty}z^k$ converges is to assume that $\displaystyle \displaystyle \lim_{m\to\infty}e^{i(m+1)\theta}$ 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 $\displaystyle \displaystyle |r| \geq 1$.

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


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

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

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

    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 $\displaystyle \displaystyle n$ terms is $\displaystyle \displaystyle \frac{a\left(r^n - 1\right)}{r - 1}$.

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

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