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Non-commutativity of a limiting sum

  1. Jul 16, 2010 #1
    I read this in the appendix of Hardy's pure mathematics textbook and an explanation was not given:

    [tex]\lim_{x \to 1} \sum_1^\infty \left( x^{n-1}-x^n \right) = \lim_{x \to 1} \left( 1 - x + (x - x^2) + \ldots \right) = \lim_{x \to 1} 1 = 1[/tex]

    [tex]\sum_1^\infty \lim_{x \to 1} \left( x^{n-1}-x^n \right) = \sum_1^\infty \left( 1 - 1 \right) = 0 + 0 + 0 + \ldots = 0[/tex]

    Are both answers correct? I thought that associativity only applied to a finite series of operations, which is why I think the first equation is wrong.
     
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  3. Jul 16, 2010 #2

    disregardthat

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    What's going on when you are taking the limit in the first equation? It seems to me that it also should be 0.

    Note however that the infinite sums only make sense when |x| <= 1.

    Remember that [tex]\lim_{x \to a } \sum^\infty_{n=1} f_n(x)[/tex] and [tex]\sum^\infty_{n=1} \lim_{x \to a } f_n(x)[/tex] are essentially two different things, and are only equal under certain conditions.
     
    Last edited: Jul 16, 2010
  4. Jul 16, 2010 #3
    [tex]\lim_{x \to 1} \left( 1 - x + (x - x^2) + \ldots \right) = \lim_{x \to 1} \left( 1 + (-x + x) + (-x^2 + x^2) + \ldots \right) = \lim_{x \to 1} \left( 1 + 0 + 0 + \ldots \right) = 1[/tex]

    This is what I meant by my associativity remark.
    In the limits, x approaches 1, meaning both [itex]\lim_{x \to 1^+}[/itex] and [itex]\lim_{x \to 1^-}[/itex]. Surely at least the "from below" limit is meets the |x| <= 1 condition?
     
  5. Jul 16, 2010 #4

    disregardthat

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    Note that [tex]\left( 1 - x + (x - x^2) + \ldots \right)[/tex] and [tex]\left( 1 + (-x + x) + (-x^2 + x^2) + \ldots \right)[/tex] are very different sums! They just happen to converge to the same value under certain conditions. That does not mean they necessarily behave equally. It must be clear what the consecutive terms are in the sum for it to make sense.


    Yes, but my point was that in the first case you are taking the limit of a sum which requires that |x| <= 1 in the first place.
     
    Last edited: Jul 16, 2010
  6. Jul 16, 2010 #5

    Office_Shredder

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    To avoid confusion: assume the limit has been removed (it doesn't exist) and instead we are just evaluating the series at x=1 (which does exist)

    For infinite sums, associativity is sketchy. In particular, you can only do infinitely many re-arrangements if your sum converges absolutely, otherwise you might be changing the value.

    For this particular example, the formal definition of an infinite series is being obscured to confuse you. Formally, an infinite series is a sequence of partial sums

    [tex]S_1,S_2,S_3,...[/tex] where [tex]S_i[/tex] is the sum of the first [tex]i[/tex] terms. The series converges to whatever number the partial sums converge to

    Writing [tex]\sum_1^\infty \left( x^{n-1}-x^n \right[/tex] means that [tex]S_1=1-x[/tex] and [tex]S_2=(1-x)+(x-x^2)[/tex] and [tex]S_3=(1-x)+(x-x^2)+(x^2-x^3)[/tex]. In general, [tex]S_n=1-x^n[/tex] and this converges to 0 as x goes to 1

    The series being used to get a value of 1 (which is not properly written out) is
    [tex]S_1=1[/tex] and [tex]S_2=1 + (-x + x)[/tex] and [tex]S_3= 1 + (-x + x) + (-x^2+x^2)[/tex] which of course has [tex]S_k=1[/tex] for all values of k.

    To summarize, these are two entirely different infinite series. What makes them look similar is the fact that they can be disguised as being the same by looking at a third divergent series
    [tex]S_1=1[/tex] and [tex]S_2=1-x[/tex] and [tex]S_3=1-x+x[/tex] and [tex]S_4=1-x+x-x^2[/tex] etc. Basically what has happened here is that there is a natural divergent series, and two different subsequences converging to different values were picked out of it. Because we're not very good intuitively at noticing this slight of hand, we assume that the two subsequences should converge to the same value, and are surprised when it doesn't


    As a final aside, you should always be careful about whether you can really move a limit from outside of an infinite series to inside of it.
     
  7. Jul 16, 2010 #6
    Wow, Office_Shredder -- that was exceptionally well put. I understand completely! Thanks a lot!
     
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