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Thread: Sum of a sequence

  1. #1
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    Sum of a sequence

    $\displaystyle \sum_{x=0}^{\infty}\frac{x}{2^x}$

    All I know is that it converges. Really have nothing more to say.
    Last edited by lausing; Feb 21st 2010 at 09:51 PM. Reason: latex
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  2. #2
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    Hello, lausing!

    $\displaystyle S \;=\;\sum^{\infty}_{x=1} \frac{x}{2^x}$

    $\displaystyle \begin{array}{cccccc}\text{We have:} & S &=& \dfrac{1}{2} + \dfrac{2}{2^2} + \dfrac{3}{2^3} + \dfrac{4}{2^4} + \hdots \\ \\[-3mm]
    \text{Multiply by }\frac{1}{2}\!: & \frac{1}{2}S &=& \quad\;\;\dfrac{1}{2^2} + \dfrac{2}{2^3} + \dfrac{3}{2^4} + \hdots \end{array}$


    . . $\displaystyle \text{Subtract: }\;\;\tfrac{1}{2}S \;\;=\;\;\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \hdots $ .[1]


    The right side is a geometric series with: .$\displaystyle a = \tfrac{1}{2},\;\;r = \tfrac{1}{2}$

    . . Its sum is: .$\displaystyle \frac{\frac{1}{2}}{1-\frac{1}{2}} \:=\:1$


    Hence [1] becomes: .$\displaystyle \tfrac{1}{2}S \;=\;1$


    Therefore: .$\displaystyle S \;=\;2$

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  3. #3
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    Thanks!

    Also, the question is erroneously posted in calculus rather than linear algebra or something, since it orginally was about derivatives. Whoops!
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