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Summation Proof Help

  1. Sep 16, 2009 #1
    [tex]\sum[/tex][tex]\frac{1}{(2j-1)^2}[/tex]

    This fgoes from j=1 to infinity. I was just wondering if somebody could calculate and show all working to show the value that this function converges to as i have no idea of how to do this? Thanks for your help
     
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  3. Sep 16, 2009 #2

    Dick

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    No one is just going to work it out for you with no input from you. Do you know how to sum 1/j^2?
     
  4. Sep 16, 2009 #3
    I never said i wasnt going to input, im happy to put input in, i just dont know how to start it. And the sum from 1/j^2 is 2 using geometric sequences. Where do i go from here tho?
     
  5. Sep 16, 2009 #4

    Dick

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    The sum of 1/j^2 isn't 2 and it isn't geometric. The sum is pi^2/6. Proving that is tricky, you have to use complex analysis or worse. My guess is that you were given that and you are expected to use that to find the sum 1/(2j-1)^2. Hint: those are the odd terms in the series 1/1^2+1/2^2+1/3^2+1/4^2+... Can you find the sum of the even terms?
     
  6. Sep 16, 2009 #5
    Just a q, is doing the sum 1/j^2 gonna help in the sum i want to calculate. Also, what do you mean by even terms as i used it when j=1,2,3, what do you specifically mean by even terms? Thanks for ur help so far, i know complex numbers and all that so i could always use complex analysis but is it easier this way, or how would i start it using complex analysis>?
     
  7. Sep 16, 2009 #6

    Dick

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    I wouldn't try computing sum 1/j^2 from scratch, it's too hard. The value is zeta(2)=pi^2/6 (where zeta is the Riemann zeta function). If you want some samples of proofs google for "zeta(2)" or "sum of inverse squares". On the other hand figuring sum 1/(2j-1)^2 is not that hard if you already know sum 1/j^2. What I mean by even and odd is split the 1/j^2 series into terms where j is even and j is odd. Your sum 1/(2j-1)^2 is the odd ones.
     
  8. Sep 16, 2009 #7
    Well, i know the odd series is 1+1/9+1/25+.... but what process would you add them all together, like what method would you use?
     
  9. Sep 16, 2009 #8

    Dick

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    If you add the sum of the odds and the sum of the evens, you get pi^2/6. Can you figure out the sum of the evens??
     
  10. Sep 16, 2009 #9
    The thing is i dont know how to add them because am i not supposed to use geometric sequence cause it isnt one. The sequence for evens is 1/4+1/16+1/36...
     
  11. Sep 16, 2009 #10

    Dick

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    Factor 1/4 out of all those terms. Does what's left over look familiar??
     
  12. Sep 16, 2009 #11
    werll 1/4 sum 1/j^2 for odds numbers equals the sum 1/j^2 for all numbers
     
  13. Sep 16, 2009 #12

    Dick

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    You mean evens, right? I'm hoping you also mean (1/4)*sum of all=sum of evens. So if sum 1/j^2 for all numbers is pi^2/6, what is that? Can you use that to deduce the sum of the odds?
     
  14. Sep 16, 2009 #13
    I got pi^2/8, so therefore, sum of j=1 to infinity of 1/(2j-1)^2 is basically pi^2/8 because theyre are equal, this is correct isnt it?
     
    Last edited: Sep 16, 2009
  15. Sep 16, 2009 #14

    Dick

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    That's not a very clear description of what you did, but yes, the sum of the odds is pi^2/8.
     
  16. Sep 16, 2009 #15
  17. Sep 16, 2009 #16

    Dick

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    That looks very nice.
     
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