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Thread: define radi of convergence and give example

  1. #1
    Senior Member
    Feb 2008

    define radi of convergence and give example

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  2. #2
    Global Moderator

    Nov 2005
    New York City
    1) Let $\displaystyle \{ a_n\}$ be be complex sequence. Say $\displaystyle w\not = 0$ is such that $\displaystyle \{ |a_nw^n| : n\geq 0\}$ is bounded. Let $\displaystyle A$ be an upper bound, so, $\displaystyle |a_nw^n| < A$ for all $\displaystyle n\geq 0$. Let $\displaystyle |z| < |w|$. Note, $\displaystyle |a_nz^n| \leq |a_nw^n| \cdot |\tfrac{z}{w} |^n < A \left| \tfrac{z}{w} \right|^n $. But $\displaystyle \left|\tfrac{z}{w}\right|^n$ is a geometric series with $\displaystyle |z/w|<1$ so it converges. The radius of convergence is define to be $\displaystyle R$ ($\displaystyle R=0$ or $\displaystyle R=\infty$ are included) so that if $\displaystyle |z|<R$ then we have convergence and if $\displaystyle |z|>R$ then we have divergence.

    2)If $\displaystyle \lim |a_n|^{1/n} = R$ then it means $\displaystyle \lim |a_n|^{1/n} |z| = R|z|$. Thus, if $\displaystyle |z| < 1/R$ then $\displaystyle R|z| < 1$. Pick $\displaystyle \epsilon > 0$ to that $\displaystyle R |z| + \epsilon < 1$. Then for sufficiently large $\displaystyle n$ we have that $\displaystyle ||a_n|^{1/n}|z| - R|z|| < \epsilon \implies R|z| - \epsilon< |a_n|^{1/n} |z| < R|z|+\epsilon$. Thus, $\displaystyle |a_nz^n| < (R|z|+\epsilon)^n$ but $\displaystyle 0<R|z|+\epsilon < 1$ so the geometric series converges. By direct comparison test it follows that $\displaystyle |a_nz^n|$ converges for $\displaystyle |z|<1/R$. Similarly if $\displaystyle |z|>1/R$ then $\displaystyle \lim |a_n|^{1/n}|z| = R|z| > 1$. Pick $\displaystyle \epsilon > 0$ so that $\displaystyle R|z| - \epsilon > 1$ and so as in above $\displaystyle |a_nz^n| > (R|z| - \epsilon)^n$ and this is a divergent geometric series.

    3)For (a) consider $\displaystyle \sum_{n=0}^{\infty} \frac{z^n}{n^2}$ using #2 the radius of convergence is $\displaystyle 1$. If $\displaystyle z$ is any point on the circle then $\displaystyle \left| \frac{z^n}{n^2} \right| \leq \frac{1}{n^2}$ so it converges too. For (b) consider $\displaystyle \sum_{n=0}^{\infty} z^n$, it does not converge on the circle because if $\displaystyle |z| = 1$ then $\displaystyle \lim |z^n| = 1\not = 0$, so by the divergence test it cannot converge. For (c) consider $\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^nz^n}{n}$. At $\displaystyle z=1$ it converges but not at $\displaystyle z=-1$.
    Last edited by ThePerfectHacker; May 27th 2008 at 03:21 PM.
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