Assumptions? Quasi-proof?

Can we assume that if $\displaystyle \sum_{n=0}^\infty a_nx^n=\sum_{n=0}^\infty b_nx^n$ for all values of $\displaystyle |x|<1$, then $\displaystyle a_n=b_n$ for all $\displaystyle n$? Somehow I don't think this is valid, but let's run with it for a moment.

If this is true, then the whole point of the two-part exercise was to make the connection that $\displaystyle \frac{1}{19!}+\frac{1}{19}+\frac{r}{s}=1$ for some $\displaystyle r,s$ where $\displaystyle 19\not|s$

Multiplying gives us $\displaystyle 1+18!=19!-19!\frac{r}{s}$ . Now reduce the $\displaystyle \frac{r}{s}$ term and because the LHS is a whole number, the RHS must reduce fully. Since $\displaystyle 19\not|s$, 19 will still be a factor in this term after reducing, so it can be written $\displaystyle 1+18!=19k$ for some positive integer k, which implies your hypothesis.

Not sure if this quasi-proof holds water, but it is the only connection I see between the two parts of the question.

Infinitely Differentiable

I think I worked it out in my head now. One of the stipulations of the proof is that the function expressible by the series is *infinitely differentiable* in the interval of convergence. So defining $\displaystyle h(x)=f(x)-g(x)$, $\displaystyle h$ could not possibly be equal to zero within some interval $\displaystyle x\in(a-\epsilon, a+\epsilon)$ and nonzero at a point outside this interval. Otherwise the nth derivative of h at $\displaystyle a+\epsilon$ would have a left hand limit of zero and a nonzero right hand limit for some $\displaystyle n$.

*Not trying to argue with mathematical proof, I'm just trying to get my intuition to catch up to it. Lemma: No infinitely differentiable function can be constant on an interval no matter how small, unless it is constant everywhere.