# Homework Help: 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 $$x,y \in \mathbb{R}$$.