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Math Help - radius of convergence of the power series

  1. #1
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    radius of convergence of the power series

    Hi, can anyone please help me with this question?

    Let \displaystyle \sum_{n=0}^{\infty} a_n z^n be a power series with complex coefficients a_n such that \displaystyle \sum_{n=0}^{\infty} \mid a_n \mid converges but \displaystyle \sum_{n=o}^{\infty} n \mid a_n \mid diverges. Prove that the radius of convergence of the power series is 1.

    I had some hint that I need to try to prove it by contradiction, first prove R \geq 1, then prove R can't be great than 1 by contradiction. But I'm not sure how to do it. Can anyone please help me? Thanks a lot.
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  2. #2
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    Recall that the radius of convergence is the same for |a_n z^n| as it is for a_nz^n. Then that R\geq 1 follows immediately from the hypotheses. Now suppose towards a contradiction that R>1, and choose m\in(1,R). Then \sum |a_n m^n| converges.

    Notice that \lim_{n\to\infty}\frac{(\ln n)/\ln m}{n}=\lim_{n\to\infty}\frac{1}{n\ln m}=0, so we can choose N such that n>\ln n/\ln m and therefore m^n>n for all n\geq N.

    It follows that \sum_{n=N}^\infty |a_n m^n|\geq\sum_{n=N}^\infty n|a_n|=\infty, a contradiction.
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  3. #3
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    Quote Originally Posted by hatsoff View Post
    Recall that the radius of convergence is the same for |a_n z^n| as it is for a_nz^n. Then that R\geq 1 follows immediately from the hypotheses. Now suppose towards a contradiction that R>1, and choose m\in(1,R). Then \sum |a_n m^n| converges.

    Notice that \lim_{n\to\infty}\frac{(\ln n)/\ln m}{n}=\lim_{n\to\infty}\frac{1}{n\ln m}=0, so we can choose N such that n>\ln n/\ln m and therefore m^n>n for all n\geq N

    It follows that \sum_{n=N}^\infty |a_n m^n|\geq\sum_{n=N}^\infty n|a_n|=\infty, a contradiction.



    Hi, thanks a lot for your reply. However, I just wonder, how you get that R \geq 1? Because the question gives series of |a_n| converges, how do I connect that with |a_n z^n| ? Thanks.
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