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Math Help - Series Convergence

  1. #1
    Junior Member Kanwar245's Avatar
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    Series Convergence

    So here is the series

    \Sigma\limits_{n=1}^\infty \frac{x^{2n-1}}{(2n-1)!}

    For what values of x will it converge?
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    Re: Series Convergence

    According to the ratio test, a series is convergent when \displaystyle \begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \end{align*}.

    So get an expression for this limit, set it less than 1, then solve for x.
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    Junior Member Kanwar245's Avatar
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    Re: Series Convergence

    Right I was thinking along the lines of ratio test

    I get this

    \lim_{n\rightarrow\infty} \frac{x^2}{2n(2n+1)}

    which becomes
    x^2\lim_{n\rightarrow\infty} \frac{1}{2n(2n+1)}

    clearly, the limit goes to 0 which is less than 1 so it converges. What can i say about x?
    Last edited by Kanwar245; March 16th 2013 at 08:21 PM.
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    Re: Series Convergence

    Since the limit is always 0 no matter the value of x, the series is convergent for all x.
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    Junior Member Kanwar245's Avatar
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    Re: Series Convergence

    ahh I see, I had that feeling too, wanted to be sure
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    Junior Member Kanwar245's Avatar
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    Re: Series Convergence

    For this series,

    \Sigma\limits_{n=1}^\infty \frac{n(x-1)^n}{2^n}

    Clearly by ratio test, I get that

    \lim_{n\rightarrow\infty}{a_{n}}^{1/n} = 1. So it's inconclusive here. What can I say about which values of x will it converge?
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    Re: Series Convergence

    Quote Originally Posted by Kanwar245 View Post
    For this series,

    \Sigma\limits_{n=1}^\infty \frac{n(x-1)^n}{2^n}

    Clearly by ratio test, I get that

    \lim_{n\rightarrow\infty}{a_{n}}^{1/n} = 1. So it's inconclusive here. What can I say about which values of x will it converge?
    If the ratio test is inconclusive you need to use a different test. Maybe the root test?
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    Junior Member Kanwar245's Avatar
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    Re: Series Convergence

    My bad, I meant the root test failed.

    EDIT: Ratio test also gives 1
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    Re: Series Convergence

    Quote Originally Posted by Kanwar245 View Post
    For this series,

    \Sigma\limits_{n=1}^\infty \frac{n(x-1)^n}{2^n}

    Clearly by ratio test, I get that

    \lim_{n\rightarrow\infty}{a_{n}}^{1/n} = 1. So it's inconclusive here. What can I say about which values of x will it converge?
    Maybe you should try the ratio test then...

    \displaystyle \begin{align*} \left| \frac{a_{n+1}}{a_n} \right| &= \left| \frac{\frac{(n+1)(x - 1)^{n+1}}{2^{n+1}}}{\frac{n(x-1)^n}{2^n}} \right| \\ &= \left| \frac{2^n(n+1)(x-1)^{n+1}}{2^{n+1}n(x-1)^n} \right| \\ &= \frac{1}{2}|x - 1| \left| \frac{n + 1}{n} \right| \end{align*}

    The "n" stuff goes to 1, but the rest doesn't. You should be able to get an interval for x when you set this entire limit less than 1.
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    Junior Member Kanwar245's Avatar
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    Re: Series Convergence

    Quote Originally Posted by Prove It View Post
    According to the ratio test, a series is convergent when \displaystyle \begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \end{align*}.

    So get an expression for this limit, set it less than 1, then solve for x.
    In order to apply the ratio test, I need to be sure that each term in the sequence is positive right, how would I know that without knowing x?
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  11. #11
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    Re: Series Convergence

    Quote Originally Posted by Kanwar245 View Post
    In order to apply the ratio test, I need to be sure that each term in the sequence is positive right, how would I know that without knowing x?
    Not true at all. For any series, if you can show the series of absolute values is convergent, then so is the original series. What we are doing by taking the absolute value of the ratio in the ratio test is to check the convergence of the absolute value series. When this is convergent, so is the original series.
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