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Power series limit proof

  1. Aug 13, 2009 #1
    1. The problem statement, all variables and given/known data

    If a_v > 0 and [tex] \sum_{v=0}^{\infty} a_v [/tex] converges, then prove that [tex] \lim_{x \rightarrow 1^-} \sum_{v=0}^{\infty} a_v x^v = \sum_{v=0}^{\infty} a_v [/tex].



    2. Relevant equations



    3. The attempt at a solution

    Since [tex] \sum a_v [/tex] converges, then we can say that [tex] \sum a_v x^v [/tex] converges for x = 1. This implies that it converges uniformly for
    all |x| < 1.

    Let [tex] f(x) = \sum a_v x^v [/tex] for |x| < 1. Since the convergence is uniform, we know that f is continuous in its interval of convergence. Since the power series converges as well for x = 1, we can also say that f(x) is "left-hand continuous" at x = 1. That is, for all positive epsilon there exists a positive delta such that [tex] |f(x) - f(1)| < \epsilon [/tex] whenever [tex] |x - 1| < \delta [/tex], if we only take x approaching 1 from the left.

    Note that [tex] |f(x) - f(1)| = |\sum a_v x^v - \sum a_v | < \epsilon [/tex] whenever [tex] |x-1| < \delta [/tex].

    This is precisely the statement: [tex] \lim_{x \rightarrow 1^-} \sum_{v=0}^{\infty} a_v x^v = \sum_{v=0}^{\infty} a_v [/tex]. QED.

    The proof seems bullet-proof to me, but what bugs me is that I didn't use the fact that the a_v are positive, so there must be something wrong with the proof...
     
  2. jcsd
  3. Aug 14, 2009 #2
    What theorem are you using to justify this (i.e., what is its exact statement)?

    I don't see how this problem can be solved without quoting Abel's Theorem, or essentially reproducing its proof. (Also, by Abel's Thm, a_v>0 is not required.)
     
  4. Aug 14, 2009 #3
    Yeah I realized the mistake I made. I have a theorem in my book stating the uniform convergence for |x| < 1, but I can't say that it is uniformly continuous at 1, I would have to use Abel's theorem. But I found another way to solve the problem. Thanks


    EDIT: The exact statement of the theorem I'm using: "If a power series in x converges for a value x = c, it converges absolutely for every value x such that |x| < |c|, and the convergence is uniform in every interval |x| <= N, where N is any positive number less than |c|. Here N may lie as near to |c| as we please.

    Just a quick question, do we ever talk about uniform convergence in open intervals? Or must we mark off some end points so that we can take off a closed interval?
     
    Last edited: Aug 14, 2009
  5. Aug 14, 2009 #4
    An example of a function with uniform convergence on an open interval (i.e. all of [tex]\mathbb{R}[/tex]) is [tex]\sin(x)[/tex]. This can be easily seen once you know the inequality [tex]|\sin x - \sin y| \leq |x - y|[/tex] for all [tex]x,y \in \mathbb{R}[/tex].
     
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