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Power series, radius of convergence and Abel's Theorem

  1. Aug 20, 2009 #1
    1. The problem statement, all variables and given/known data
    Suppose the series [tex]\sum_{n=0}^{\infty} a_n x^n[/tex] has radius of convergence [tex]R[/tex] and converges at [tex]x = R[/tex]. Prove that [tex]\lim_{x \to R^{-}}\large( \sum_{n = 0}^{\infty} a_n x^n \large) = \sum_{n = 0}^{\infty} \large( \lim_{x \to R^{-}} a_n x^n \large) [/tex]


    2. Question
    For the case [tex]R \in \mathbb{R}-\{0\}[/tex] (i.e. the radius of convergence is a finite nonzero real number), then the above is a straight forward application of Abel's Theorem on the LHS, and an easy application of the fact that [tex]x^n[/tex] is a continuous function on [tex]\mathbb{R}[/tex] on the RHS. If [tex]R = 0[/tex], then the result is trivial.

    However, my question is how do you (or do you even) deal with the case [tex]R = \pm \infty[/tex]? That is, do we even say that the power series converges at [tex]x = R = \pm \infty[/tex]?

    Thanks in advance!
     
  2. jcsd
  3. Aug 20, 2009 #2
    No. [tex] \pm \infty [/tex] isn't a real number.

    Say x = x' is an interior point in the interval of convergence (we can assume x' is positive without any loss of generality), then the series converges uniformly for x' and the theorem you proved follows trivially from Abel's theorem. But if [tex] R = \infty [/tex] then you can always find a value x'' > x' so that x' is an interior point in the interval [0, x''] and so the series converges uniformly for x' and so your theorem follows again for x'.
     
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