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Math Help - [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  \abs{r} <1that x < \frac{1}{2}
    Last edited by xwrathbringerx; January 29th 2010 at 10:06 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  \abs{r} <1that x < \frac{1}{2}
    Writing this series in sigma notation we have

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


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


    The limiting sum of a geometric series is

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

    In this case, this is

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


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

    So |-2x| < 1

    -1 < -2x < 1

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

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

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




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


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

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

    3 = 2(1 + 2x)

    3 = 2 + 4x

    1 = 4x

    x = \frac{1}{4}.
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