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Odd series behavior?

  1. Mar 13, 2009 #1
    Today, my professor said something like "The series 1 + -1 + 1 + -1 and so on is defined to be one half... but let's not go into that." and then didn't feel like explaining when people asked him why. I have no idea why that would be true...

    It seems like a similar case might be

    [tex]\int_{0}^{\infty}\sin x\,\textrm{d}x[/tex]

    but that isn't defined to be one half or zero or anything at all.

    So why oh why is this true?

  2. jcsd
  3. Mar 13, 2009 #2
    THis is true.....because it is NOT true!

    the alternating series

    does not converge at all.
    One will get either 1 or 0 as the final sum, depending on how you group the terms. So, by the definition of what we mean with convergence it does not converge.

    BUT, i have heard of some sort of Eulers method, or Ramaujan summation, or some different kind of summation, and that might be true, but i have no knowledge whatsoever of those summations.

    So, this is not true, if the summation is the common one, i don't know about the others. But people here will enlighten you, just wait until this thread catches the eyes of the right people....
  4. Mar 14, 2009 #3
  5. Mar 14, 2009 #4
    Another reason why one might want the sum to be 1/2 is that


    for [tex]-1<x<1[/tex]. If you put in [tex]x=-1[/tex], then the series on the left diverges, but on the right you get 1/2. This is basically the idea behind "Abel summation", which is a weaker form of Cesaro's method.
  6. Mar 14, 2009 #5
    I've usually seen Ramanujan summation being used in cases like that.

    Equivalently you can write it in terms of the zeta function, by
    [tex]\eta(s)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s} = (1-2^{1-s})\zeta(s)[/tex]
    so that,

    [tex]\sum_{n=0}^\infty (-1)^{n} = \eta(0) = -\zeta(0) = \frac{1}{2}[/tex]

    If you have seen the derivation of the functional equation (which is as far as I have understood is Ramanjuan summation) for the zeta function this sort of "makes sense".
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