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Thread: [SOLVED] Limiting Sum of Geometric Series

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    Exclamation [SOLVED] Limiting Sum of Geometric Series

    For what range of values of x will the series 1-2x+4x^2-8x^3+... have a limting sum? What is the value of x if the limting sum is 2/3?

    Is it because $\displaystyle \abs{r} <1that x < \frac{1}{2}$
    Last edited by xwrathbringerx; Jan 29th 2010 at 09:06 PM.
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    Quote Originally Posted by xwrathbringerx View Post
    For what range of values of x will the series 1-2x+4x^2-8x^3+... have a limting sum? What is the value of x if the limting sum is 2/3?

    Is it because $\displaystyle \abs{r} <1that x < \frac{1}{2}$
    Writing this series in sigma notation we have

    $\displaystyle 1 - 2x + 4x^2 - 8x^3 + \dots = \sum_{i = 0}^{\infty}(-2x)^i$.


    This is a geometric series with $\displaystyle a = 1$ and $\displaystyle r = -2x$.


    The limiting sum of a geometric series is

    $\displaystyle S_{\infty} = \frac{a}{1 - r}$ provided that $\displaystyle |r| < 1$

    In this case, this is

    $\displaystyle \frac{1}{1 - (-2x)} = \frac{1}{1 + 2x}$.


    But of course, this is provided that $\displaystyle |r| < 1$

    So $\displaystyle |-2x| < 1$

    $\displaystyle -1 < -2x < 1$

    $\displaystyle \frac{1}{2} > x > -\frac{1}{2}$

    $\displaystyle -\frac{1}{2} < x < \frac{1}{2}$

    $\displaystyle |x| < \frac{1}{2}$.




    So the limiting sum is $\displaystyle \frac{1}{1 + 2x}$ for $\displaystyle |x| < \frac{1}{2}$.


    If the limiting sum is $\displaystyle \frac{2}{3}$ then

    $\displaystyle \frac{1}{1 + 2x} = \frac{2}{3}$

    $\displaystyle 3 = 2(1 + 2x)$

    $\displaystyle 3 = 2 + 4x$

    $\displaystyle 1 = 4x$

    $\displaystyle x = \frac{1}{4}$.
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