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Adding Series, What is wrong here?

  1. Sep 21, 2012 #1
    1. The problem statement, all variables and given/known data
    With binomial expansions,
    [itex]\frac{x}{1-x}[/itex] = [itex]\sum[/itex][itex]^{n=\infty}_{n=1}[/itex]xn
    [itex]\frac{x}{x-1}[/itex] = [itex]\sum[/itex][itex]^{n=\infty}_{n=0}[/itex]x-n

    Adding these series yields:

    [itex]\sum[/itex][itex]^{n=\infty}_{n=-\infty}[/itex]xn=0
    This is nonsense, but what went wrong here?

    3. The attempt at a solution
    Obviously, [itex]\frac{x}{1-x}[/itex]+[itex]\frac{x}{x-1}[/itex]=0 (1)

    It's clear that they tried to transform [itex]\frac{x}{x-1}[/itex] = [itex]\sum[/itex][itex]^{n=\infty}_{n=0}[/itex]x-n into [itex]\frac{x}{x-1}[/itex] = [itex]\sum[/itex][itex]^{n=0}_{n=-\infty}[/itex]xn and then substitute into equation (1) and get an answer of zero.

    From there, I'm not sure how in the world they manipulated that series to get a neg. infinity to show up in the limits, and where the mistake in that is.
     
  2. jcsd
  3. Sep 21, 2012 #2

    HallsofIvy

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    The second is wrong. x- 1= -(1- x) so the second series is just
    [itex]\sum_{n=1}^\infty -x^n[/itex], not [itex]\sum_{n=1}^\infty x^{-n}[/itex].

     
  4. Sep 21, 2012 #3

    jbunniii

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    The first series converges only if |x| < 1. The second converges only if |x| > 1. Therefore there is no x for which you can add the series and get a meaningful result.
     
  5. Sep 21, 2012 #4
    I don't understand this. I realize this myself, and if I were deriving the series myself, I'd use that technique and get the same result you. Also, when I use Maple to get the series, the result also matches yours. But, reference tables I have on hand show series matching what was printed in the homework.
     
  6. Sep 21, 2012 #5

    jbunniii

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    If |x| > 1, then
    [tex]\sum_{n = 0}^{\infty} x^{-n} = \frac{1}{1 - x^{-1}} = \frac{x}{x - 1}[/tex]
    so the series expansion is correct. The problem is that the region of convergence is incompatible with that of the first series.
     
  7. Sep 21, 2012 #6

    Ray Vickson

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    If you expand x/(1-x) as a series in t = 1/x you will get the printed result. One expansion applies for small |x| and the other applies for large |x|.

    RGV
     
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