Let $f:\mathbb{R} \rightarrow \mathbb{R}$ and let f be a smooth function. Let R denote the radius of convergence of the Taylor series, centered at a, of f. For each $n \in \mathbb{N}$, let $M_n = \sup \{|f^{(n)} (t)|: t \in \mathbb{R} \}$. $$ \mbox{If}\quad\lim_{n\rightarrow \infty} \frac{M_n}{n!} R^{n} < \infty \quad\mbox{then}\quad f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)} (a)}{n!} (x-a)^n \forall x \in (a - R, a + R) $$
A hint was given saying to show that $$\mbox{if}\quad x \in (a - R, a + R) \quad\mbox{then}\quad \lim_{n\rightarrow \infty} \frac{M_n}{n!} (x-a)^n = 0 $$
Edit: I think I just figured out why the hint is helpful. Below is my proposed proof. A corollary to Taylor's theorem says that the error term is bounded by:
$|f(x) - \sum \frac{1}{n!} f^{n} (c) (x-c)^n \leq | \frac{|x-c|}{(k+1)!} \sup | f(t)^{k+1}|$
So I'm guessing that if we show that $lim_{n\rightarrow \infty} \frac{M_n}{n!}R^{n}=0$, then we're just left with $f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)} (a)}{n!} (x-a)^n$, which would exist by definition since f is a smooth function (n times differentiable for all $\mathbb{R}$). But I'm still not sure how I'm supposed to prove the hint.
EDIT: Ok, I know that if $\sum \frac{M_n}{n!} (x-a)^n = 0$, then $\lim \frac{M_n}{n!} (x-a)^n=0$. So I'm thinking of using the radius of convergence of power series theorem. And so I know that we will have convergence if $|x-a| < \underline{\lim} ( \frac{M_n}{n!} )^{1/n}$. I have no idea on how to prove this, or if this is even the correct method. Help would be much appreciated.