Power series limit proof

1. Aug 13, 2009

JG89

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. The attempt at a solution

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

Let $$f(x) = \sum a_v x^v$$ 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 $$|f(x) - f(1)| < \epsilon$$ whenever $$|x - 1| < \delta$$, if we only take x approaching 1 from the left.

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

This is precisely the statement: $$\lim_{x \rightarrow 1^-} \sum_{v=0}^{\infty} a_v x^v = \sum_{v=0}^{\infty} a_v$$. 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. Aug 14, 2009

Billy Bob

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.)

3. Aug 14, 2009

JG89

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
4. Aug 14, 2009

fmam3

An example of a function with uniform convergence on an open interval (i.e. all of $$\mathbb{R}$$) is $$\sin(x)$$. This can be easily seen once you know the inequality $$|\sin x - \sin y| \leq |x - y|$$ for all $$x,y \in \mathbb{R}$$.