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

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

    3)For (a) consider \sum_{n=0}^{\infty} \frac{z^n}{n^2} using #2 the radius of convergence is 1. If z is any point on the circle then \left| \frac{z^n}{n^2} \right| \leq \frac{1}{n^2} so it converges too. For (b) consider \sum_{n=0}^{\infty} z^n, it does not converge on the circle because if |z| = 1 then \lim |z^n| = 1\not = 0, so by the divergence test it cannot converge. For (c) consider \sum_{n=0}^{\infty} \frac{(-1)^nz^n}{n}. At z=1 it converges but not at z=-1.
    Last edited by ThePerfectHacker; May 27th 2008 at 04:21 PM.
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