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Help me find this infinite summation

  1. Jul 30, 2006 #1
    [tex]\sum_{x =1}^{\infty} x (\frac{1}{2})^x[/tex]

    is there a formula for this, like for infinte geometric summation?
     
    Last edited: Jul 30, 2006
  2. jcsd
  3. Jul 30, 2006 #2

    matt grime

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    Yes. Given x(1/2)^x, can you think of some other quantity that you can apply something well known to to get x(1/2)^x, or perhaps more suggestively x(1/2)^{x-1}?
     
  4. Jul 30, 2006 #3

    0rthodontist

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    Explaining that a little, consider it as a special case of
    [tex]\sum_{x =1}^{\infty} x y^x[/tex]

    If you are familiar with generating functions for sequences, multiplying every nth term by n is a standard operation.
     
  5. Jul 30, 2006 #4
    I think, I get what you are trying to say...

    since
    [tex]\sum_{x=1}^{\infty} (\frac{1}{2})^x = 1[/tex] by the infinite geometric summation formula

    then, is it correct that:
    [tex]\sum_{x=1}^{\infty} x(\frac{1}{2})^x = x[/tex] ?
     
    Last edited: Jul 30, 2006
  6. Jul 30, 2006 #5

    StatusX

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    No. Try replacing 1/2 with a variable like y as Orthodontist suggested, and differentiate the sum with respect to y.
     
  7. Jul 30, 2006 #6
    oh, I think I got what you are trying to say:
    [tex]\sum_{x=1}^{\infty} y^x = \frac{y}{1-y}[/tex]
    differentiating, I get:
    [tex]\sum_{x=1}^{\infty}x y^{x-1} = \frac{1}{(1-y)^2}[/tex]
    multiplying by y,
    [tex]\sum_{x=1}^{\infty}x y^x = \frac{y}{(1-y)^2}[/tex]
    so,
    [tex]\sum_{x=1}^{\infty}x (\frac{1}{2})^x = 2 [/tex]

    Is this correct?
     
  8. Jul 30, 2006 #7

    0rthodontist

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    Yes it is.
     
  9. Jul 31, 2006 #8
    great! thanks for your help, matt, orthodontist, statusX
     
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