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Math Help - [SOLVED] A demonstration regarding the radius of convergence of an infinite series

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] A demonstration regarding the radius of convergence of an infinite series

    Given the power series \sum _{n=0}^{\infty} a_n (z-z_0)^n, prove that if \lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}}   \right| exists and is finite or infinite, then it is equal to the radius of convergence of the series.
    Suggestion : Use the ratio test.

    My attempt : The ratio test states that if L= \lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}}   \right| exists and if L<1 then the series converges. If L=1 then we can't conclude by the test. If L>1 then the series diverges.

    The radius of convergence of the power series is R=\frac{1}{\lim _{n \to \infty} |a_n|^{\frac{1}{n}} }.

    Thus I must show that \frac{1}{\lim _{n \to \infty} |a_n|^{\frac{1}{n}} }=\lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}}   \right|.
    I first assume that L<1, in other words that |a_n|<|a_{n+1}|.
    I've tried to continue the proof ( by showing \frac{1}{\lim _{n \to \infty} |a_n|^{\frac{1}{n}} }=\lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}}   \right| using the fact that |a_n|<|a_{n+1}| ) but without success.

    Can you help me a little bit? Thanks.
    Last edited by arbolis; October 17th 2009 at 10:10 AM.
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  2. #2
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    Opalg's Avatar
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    Quote Originally Posted by arbolis View Post
    Given the power series \sum _{n=0}^{\infty} a_n (z-z_0)^n, prove that if \lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}}   \right| exists and is finite or infinite, then it is equal to the radius of convergence of the series.
    Suggestion : Use the ratio test.

    My attempt : The ratio test states that if \color{red}L= \lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}}   \right| exists and if \color{red}L<1 then the series converges. If \color{red}L=1 then we can't conclude by the test. If \color{red}L>1 then the series diverges.

    The radius of convergence of the power series is R=\frac{1}{\lim _{n \to \infty} |a_n|^{\frac{1}{n}} }.

    Thus I must show that \frac{1}{\lim _{n \to \infty} |a_n|^{\frac{1}{n}} }=\lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}}   \right|.
    I first assume that L<1, in other words that |a_n|<|a_{n+1}|.
    I've tried to continue the proof ( by showing \frac{1}{\lim _{n \to \infty} |a_n|^{\frac{1}{n}} }=\lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}}   \right| using the fact that |a_n|<|a_{n+1}| ) but without success.

    Can you help me a little bit? Thanks.
    The sentence in red is a correct statement about the series \sum _{n=0}^{\infty} a_n . But you are trying to apply it to the series \sum _{n=0}^{\infty} a_n (z-z_0)^n, in which the n'th term is not a_n but a_n(z-z_0)^n. So what the ratio test is telling you is that if L= \lim _{n \to \infty} \left| \frac{a_n(z-z_0)^n}{a_{n+1}(z-z_0)^{n+1}} \right| exists and if L<1 then the series converges. If L=1 then we can't conclude by the test. If L>1 then the series diverges.

    But \lim _{n \to \infty} \left| \frac{a_n(z-z_0)^n}{a_{n+1}(z-z_0)^{n+1}} \right| = \frac1{|z-z_0|}\lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|. From that, you can conclude directly that the power series converges if |z-z_0|<\lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}}   \right|, and diverges if |z-z_0|>\lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}}   \right|.
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  3. #3
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Opalg View Post
    The sentence in red is a correct statement about the series \sum _{n=0}^{\infty} a_n . But you are trying to apply it to the series \sum _{n=0}^{\infty} a_n (z-z_0)^n, in which the n'th term is not a_n but a_n(z-z_0)^n. So what the ratio test is telling you is that if L= \lim _{n \to \infty} \left| \frac{a_n(z-z_0)^n}{a_{n+1}(z-z_0)^{n+1}} \right| exists and if L<1 then the series converges. If L=1 then we can't conclude by the test. If L>1 then the series diverges.

    But \lim _{n \to \infty} \left| \frac{a_n(z-z_0)^n}{a_{n+1}(z-z_0)^{n+1}} \right| = \frac1{|z-z_0|}\lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|. From that, you can conclude directly that the power series converges if |z-z_0|<\lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}}   \right|, and diverges if |z-z_0|>\lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}}   \right|.
    Oh thank you very much. That was very useful!
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