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Proof of the value of an infinite sum

  1. Sep 11, 2008 #1
    1. The problem statement, all variables and given/known data
    On an exam we were given that the [tex]\Sigma[/tex] 1/n^2 was (pi^2)/6. We were asked to prove that the value of [tex]\Sigma[/tex] ((-1)^(n+1))/(n^2)) = (pi^2)/12.

    I'm sorry about the lack of latex; this is the first time I've ever *tried* to use it.

    2. Relevant equations
    Basic properties of series.
    [tex]\Sigma[/tex] 1/n^2 = (pi^2)/6

    3. The attempt at a solution
    I rewrote the series as:
    [tex]\Sigma[/tex] 1/(2n+1)^2 - [tex]\Sigma[/tex]1/(2n)^2
    From the given information I know that [tex]\Sigma[/tex] 1/(2n)^2 = (pi^2)/24.
    The problem is that fact that I don't know how to find [tex]\Sigma[/tex] 1/(2n+1)^2.
    I used the integral test to approximate it, and I got 5/4, which is extremely close to the value it should be, which is (pi^2)/8, but I don't think this counts as "proving" it.

    Can anyone think of an alternative way to go about proving it, or finding the exact value of [tex]\Sigma[/tex] 1/(2n+1)^2?

    Last edited: Sep 11, 2008
  2. jcsd
  3. Sep 11, 2008 #2


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    The sum of 1/(2n)^2 + the sum of 1/(2n+1)^2 = the sum of 1/n^2. Isn't it?
  4. Sep 11, 2008 #3
    ...I can't believe I overlooked that. It was sitting in front of me the whole time :(.

    Thanks a ton.
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