1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    Thanks Benorin. That's very nice.:smile:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook