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Another Solvable Series

  1. Aug 21, 2004 #1
    Another "Solvable" Series

    Hi

    This is another series

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

    I did it by dividing the numerator by the denominator (and doing the division once again) and finally using partial fractions. Is there a neater way? (I will post my solution to the problem in a few hours...going out right now).

    Thanks a lot,

    Cheers
    Vivek
     
  2. jcsd
  3. Aug 21, 2004 #2

    Zurtex

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    I've started by saying that:

    [tex]\frac{i^4}{(2i-1)(2i+1)} = Ai^2 + Bi + \frac{C}{2i-1} + \frac{D}{2i + 1}[/tex]

    Once solved for A, B, C and D the sum becomes:

    [tex]S = \left( A\sum_{i = 1}^{n} i^2 \right) + \left( B\sum_{i = 1}^{n} i \right) + \left( \sum_{i = 1}^{n} \frac{C}{2i-1} + \frac{D}{2i + 1} \right)[/tex]

    The former 2 sums are standard form, the latter one is a quite simple as it takes the form [itex]f(i) = g(i) - h(i)[/itex]

    That's the simplest way I see of doing it, but someone else might come up with a better method.
     
  4. Aug 21, 2004 #3
    Hi Zurtex

    That's EXACTLY what I did :biggrin:

    Thanks anyway for clarifying my method. Yeah, lets see if someone comes up with something different.

    Cheers
    Vivek
     
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