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Pi series

  1. Dec 11, 2006 #1
    Hey guys!

    My friend showed me a infinite series on maple

    infinity
    ---
    \ ( ((n-1)^2) - (1/4) )^(-1) = (Pi/2)
    /
    ---
    n=1

    Is anyone familar with this?
    If so can you refer me to its proof? Or maybe post it.

    Thanks
     
  2. jcsd
  3. Dec 11, 2006 #2

    Hurkyl

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    Gold Member

    Hrm. Unless I made a mistake, that sums to 2. You meant

    [tex]
    \sum_{n = 1}^{+\infty} \frac{1}{ (n-1)^2 - \frac{1}{4}}
    [/tex]

    right? Anyways, the usual trick to these is to use partial fractions.
     
  4. Dec 11, 2006 #3
    Sorry i think the (1/4) is positive

    that should give (Pi/2)

    yes/no?
     
  5. Dec 11, 2006 #4
    Sorry heres the correct series

    infinity
    ----
    \ (((2*n-1)^2) - (1/4))^(-1)
    /
    ----
    n=1
     
  6. Dec 12, 2006 #5

    dextercioby

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    Use partial fractions, as hinted already.

    [tex] \frac{1}{\left(2n-1\right)^{2}-\frac{1}{4}}=2\left(\frac{1}{4n-3}-\frac{1}{4n-1}\right) [/tex]

    Daniel.
     
  7. Dec 12, 2006 #6
    That works but are there any other options?
     
    Last edited: Dec 12, 2006
  8. Dec 13, 2006 #7

    dextercioby

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    I don't see another one.

    Daniel.
     
  9. Dec 14, 2006 #8
    Im interested in the ways you can obtain the series.
     
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