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Infinite series

  1. Sep 26, 2007 #1
    1. The problem statement, all variables and given/known data
    I am wondering if someone could give me some insight on how the following infinite series was derived:

    [tex] P_e = \sum_{-\infty}^\infty (1/2)^{2|n|} = -1 + 2 \sum_{n=0}^\infty (1/2)^{2n} = 5/3 [/tex]

    2. Relevant equations
    See above

    3. The attempt at a solution
    I think the -1 comes when n = 0 and the 2 before the sum is because the absolute value of n makes the result symetrical around 0. That is why one can make the sum from 0 to infinity and multiply by 2. Right??
    The second sumation must be equal to 4/3. Right? I guess my real question then is how do you find the closed form of this infinite series?
  2. jcsd
  3. Sep 27, 2007 #2


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    Can you do this one?

    [tex] \sum_{n=0}^\infty x^n[/tex]
  4. Sep 27, 2007 #3


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    Actually, the "-1" comes from the fact that when n= 0, [itex](1/2)^{2|n|}[/itex] is equal to 1. Since you are multiplying the sum, from 0 to infinity, by 2, you are getting that twice and need to subract off one.
  5. Sep 27, 2007 #4
    Yes, that is equal to 1/(1- 1/2) or 2
  6. Sep 28, 2007 #5


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    It is if x=1/2. What is it for general x? Then, what value of x applies to your problem? (Hint: it's not 1/2.)
  7. Sep 28, 2007 #6
    general x is x^2 and my x is 1/4 so 1/(1-1/4) is 4/3. All right! Thanks!
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