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Math Help - Limit of this series?

  1. #1
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    Limit of this series?

    \sum_{n=0}^{\infty}(\frac{x}{2})^{2n+1}

    Not sure how to go abou this..

    Any help?
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  2. #2
    Super Member flyingsquirrel's Avatar
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    This series looks like a geometric series...

    \sum_{n=0}^{\infty}\left(\frac{x}{2}\right)^{2n+1}  =\frac{x}{2}\sum_{n=0}^{\infty}\left(\frac{x^2}{4}  \right)^{n}
    Does it help ?
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  3. #3
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    Aha! Thank you, I got it
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  4. #4
    Super Member Showcase_22's Avatar
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    How did you do it?

    It looks like it might involve cosh (does it?).
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  5. #5
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    How did you do it?
    \sum_{n=0}^{\infty}\left(\frac{x^2}{4}\right)^{n} is a geometric series (common ratio : x^2/4) hence

    \frac{x}{2}\sum_{n=0}^{\infty}\left(\frac{x^2}{4}\  right)^{n}=\frac{x}{2}\cdot \frac{1}{1-\frac{x^2}{4}}=\frac{2x}{4-x^2}\quad \text{\ if\ } \left|\frac{x^2}{4}\right|<1
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  6. #6
    Super Member Showcase_22's Avatar
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    ahh, i'm clearly getting a bit rusty after my exam.

    Thanks!
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  7. #7
    Moo
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    Quote Originally Posted by Showcase_22 View Post
    How did you do it?

    It looks like it might involve cosh (does it?).
    It would, if there were (2n)! in the denominator
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