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Thread: [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 $\displaystyle \sum _{n=0}^{\infty} a_n (z-z_0)^n$, prove that if $\displaystyle \lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|$ 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 $\displaystyle L= \lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|$ exists and if $\displaystyle L<1$ then the series converges. If $\displaystyle L=1$ then we can't conclude by the test. If $\displaystyle L>1$ then the series diverges.

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

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

    Can you help me a little bit? Thanks.
    Last edited by arbolis; Oct 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 $\displaystyle \sum _{n=0}^{\infty} a_n (z-z_0)^n$, prove that if $\displaystyle \lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|$ 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 $\displaystyle \color{red}L= \lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|$ exists and if $\displaystyle \color{red}L<1$ then the series converges. If $\displaystyle \color{red}L=1$ then we can't conclude by the test. If $\displaystyle \color{red}L>1$ then the series diverges.

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

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

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

    But $\displaystyle \lim _{n \to \infty} \left| \frac{a_n(z-z_0)^n}{a_{n+1}(z-z_0)^{n+1}} \right| = \frac1{|z-z_0|}\lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|$. From that, you can conclude directly that the power series converges if $\displaystyle |z-z_0|<\lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|$, and diverges if $\displaystyle |z-z_0|>\lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|$.
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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 $\displaystyle \sum _{n=0}^{\infty} a_n $. But you are trying to apply it to the series $\displaystyle \sum _{n=0}^{\infty} a_n (z-z_0)^n$, in which the n'th term is not $\displaystyle a_n$ but $\displaystyle a_n(z-z_0)^n$. So what the ratio test is telling you is that if $\displaystyle L= \lim _{n \to \infty} \left| \frac{a_n(z-z_0)^n}{a_{n+1}(z-z_0)^{n+1}} \right|$ exists and if $\displaystyle L<1$ then the series converges. If $\displaystyle L=1$ then we can't conclude by the test. If $\displaystyle L>1$ then the series diverges.

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