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How to prove Carrier's rule for divergent series?

  1. Jul 12, 2008 #1
    Carrier's Rule :confused:
     
  2. jcsd
  3. Jul 12, 2008 #2

    HallsofIvy

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    Did you see the word "humorous" in that description? There is no proof, its a joke.
     
  4. Jul 12, 2008 #3
    Yes, I know. But it is still the case that a few terms of an asymptotic series wil give you a good approximation while if you use a series that converges you typically need may terms to get a similar approximation.

    This is then alleged to be caused by the fact that demanding convergence is an extra burden that comes at the expense of how good the approximation can be after only a few terms of the series.

    Is there a general theorem about this?
     
  5. Jul 13, 2008 #4

    HallsofIvy

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    No, that is not the case. It is quite possible to have a convergent series that converges rapidly or an asymptotic series that is very slow. It's just that there is no reason to talk about asymptotic series like that. We only look at "good" asymptotic series.

     
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