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**JG89** Problem:

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

Attempt:

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

all |x| < 1.

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

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

This is precisely the statement: $\displaystyle \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...