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Double Series

  1. Nov 10, 2005 #1
    Hello, all--
    I have the following series that I'm supposed to calculate:

    [tex]\sum_{i=1}^{\infty} \sum_{j=i}^{\infty} p^(j+i)[/tex]

    Could someone show me how it should be done?

    Note: It's supposed to be p^(j+i) or p raised to j+i; however, I can't get the formatting to work.

    Last edited: Nov 10, 2005
  2. jcsd
  3. Nov 10, 2005 #2


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    Here's the formatting:

    [tex]\sum_{i=1}^{\infty} \sum_{j=i}^{\infty} p^{(j+i)}[/tex]

    Suppose p was between 0 and 1, say 1/2 and we start with i=1. The first member of the set would then be:


    second one, well you know what it it. Third one . . . They converge don't they? Suppose they all converge to numbers less than 1. I don't know but suppose they did and you were able to express the general form of those numbers say:





    Then we could sum them right? Not saying that's what they would be but this is just one way I can think of.
  4. Nov 12, 2005 #3


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    The double seires you have presented is convergent only for [itex] \left| p\right| <1[/itex] (p may be real or complex), for these values of p we have:

    [tex]\sum_{i=1}^{\infty} \sum_{j=i}^{\infty} p^{j+i} = \sum_{i=1}^{\infty} \left( p^{i} \sum_{j=i}^{\infty} p^{j}\right) = \sum_{i=1}^{\infty} \left( p^{2i} \sum_{j=i}^{\infty} p^{j-i}\right) [/tex]
    [tex]=\sum_{i=1}^{\infty} \left( p^{2i} \sum_{j=0}^{\infty} p^{j}\right) = \left( \sum_{i=1}^{\infty} p^{2i}\right) \left( \sum_{j=0}^{\infty} p^{j}\right) [/tex]
    [tex]= \left(p^{2} \sum_{i=0}^{\infty} \left( p^{2}\right)^{i}\right) \left( \frac{1}{1-p}\right) = \left(\frac{p^{2}}{1-p^{2}} \right) \left( \frac{1}{1-p}\right) = \frac{p^{2}}{(1-p)^{2}(1+p)}[/tex]
  5. Nov 12, 2005 #4


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    Thanks Benorin. That's very nice.:smile:
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