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Geometric Series with probability

  1. Mar 16, 2009 #1
    Using the formula for the sum of geometric series, show that the values of p(n) sum to 1

    p(n)=[tex](1 - \alpha)^n \alpha[/tex]

    My attempt:
    [tex]
    \alpha
    \sum^\infty_{{\bf n=0}}
    (1- \alpha)^n
    [/tex]

    I am not sure where to go from here. Any help to show this is true!
     
  2. jcsd
  3. Mar 16, 2009 #2
    Do you know the formula

    [tex]\sum_{i=0}^{\infty}x^i=\frac{1}{1-x}[/tex] valid for [tex]|x|<1[/tex] ?

    Apply it and you are almost done.
     
  4. Mar 16, 2009 #3

    Dick

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    What's the sum of the geometric series r^n? Now just put r=1-alpha.
     
  5. Mar 16, 2009 #4
    [tex]
    \alpha \sum_{i=0}^{\infty}x^i=\frac{1}{1-(1-\alpha)} -1 = \frac{(1-\alpha)}{1-(1- \alpha)}
    [/tex]

    I think I got it. Does this look right.
     
  6. Mar 16, 2009 #5

    Dick

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    Not really. Can you fix it? Sum 0 to infinity of r^n is 1/(1-r) for |r|<1 as I recall.
     
  7. Mar 17, 2009 #6
    [tex]
    \alpha \frac{1}{1-(1-\alpha)} = \frac{\alpha}{0 + \alpha}

    [/tex]
     
  8. Mar 17, 2009 #7

    Dick

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    And alpha/alpha=?
     
  9. Mar 17, 2009 #8
    1. Thanks everybody for the help on this one.
     
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