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Series, it's a riddle to me !

  1. Nov 26, 2008 #1
    Hey, guys !
    I have to calculate:

    [tex]\sum 1/(2n + 1)^2 [/tex]
    n from 0 to infinite

    Knowing that 1/n² = (pi)²/6

    The answer is (pi)²/8

    I don't know why I'm stuck here but I just don't get it...

    Greetings
     
    Last edited: Nov 26, 2008
  2. jcsd
  3. Nov 26, 2008 #2

    Office_Shredder

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    Once you have what

    [tex]\sum_{n=1}^{ \infty } \frac{1}{n^2}[/tex] is, you should be able to find what

    [tex]\sum_{n=1}^{ \infty } \frac{1}{(2n)^2}[/tex] is pretty easily. And then use both of these to find the series sum that you're looking for
     
  4. Nov 26, 2008 #3
    It is given ...
    Calculate:
    [tex]\sum^{\infty}_{0} 1/(2n +1)^2 [/tex]

    if you know that:

    [tex]\sum^{\infty}_{1} 1/n^2 = (pi)^2 /6 [/tex]


    But I can't make it work :(
     
  5. Nov 26, 2008 #4

    Office_Shredder

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    Can you calculate this?


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

    Then what is


    [tex]
    \sum_{n=1}^{ \infty } \frac{1}{(2n)^2} + \sum_{n=0}^{ \infty } \frac{1}{(2n+1)^2}
    [/tex]
     
  6. Nov 26, 2008 #5
    haha I'm sorry for wasting your time !
    It was infact not hard, but had a few drinks too much last night.

    [​IMG]


    thank you anyway !
     
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