If a_v > 0 and converges, then prove that .
Since converges, then we can say that converges for x = 1. This implies that it converges uniformly for
all |x| < 1.
Let 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 whenever , if we only take x approaching 1 from the left.
Note that whenever .
This is precisely the statement: . 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...