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Help with a series question

  1. Jan 11, 2013 #1
    This is the last part to a Fourier Series problem. After using Parseval's Identity the following series emerges: [itex]\frac{1}{1^{4}}+\frac{1}{3^{4}} +\frac{1}{5^{4}} +\frac{1}{7^{4}}+\cdots = \frac{\pi^{4}}{96}[/itex]

    We are then asked to show: [itex]\frac{1}{1^{4}}+\frac{1}{2^{4}} +\frac{1}{3^{4}} +\frac{1}{4^{4}}+\cdots = \frac{\pi^{4}}{90}[/itex]



    Must one split the left hand side of the last identity into [itex]\sum\frac{1}{(2n)^{4}} + \sum\frac{1}{(2n-1)^{4}}[/itex], or is there an easier way? I'm guessing there is.

    Thanks a lot.
     
  2. jcsd
  3. Jan 11, 2013 #2

    Curious3141

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    Hint:

    Call your original series (sum of reciprocals of powers of odds) S. You're supposed to determine ##\zeta(4)##

    ##\sum\frac{1}{(2n-1)^{4}} = S##

    ##\sum\frac{1}{(2n)^{4}} = \frac{1}{16}\zeta(4)##

    and the sum of those two is equal to ##\zeta(4)## again, right?

    Just rearrange and solve for ##\zeta(4)##.
     
  4. Jan 11, 2013 #3
    But are you assuming that ##\sum\frac{1}{(2n)^{4}} = \frac{1}{16}\zeta(4)##? Oh wait I think I understood. I'll let you know later thanks a lot.
     
  5. Jan 11, 2013 #4

    Curious3141

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    You're just bringing the constant term ##\frac{1}{2^4}## outside the summation sign.
     
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