# f is a smooth function, and $M_n$ is the sup of $f^{(n)}$. Show if $\lim_{n \to \infty} \frac{M_n}{n!}R^n < \infty$, then f(x) is the taylor series.

Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a smooth function (i.e. assume that the n-th derivative $f^{(n)}$ is defined on all of $\mathbb{R}$). Let $R$ denote the radius of convergence of the Taylor series of $f$, centered at $a$. For each $n \in N$, let $M_n = sup$ {$|f^{(n)}(t)|:t \in (a-R, a+R)$}.

Show that if $\lim_{n \to \infty} \frac{M_n}{n!}R^n < \infty$, then $f(x) = \sum\limits_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n$ for every $x \in (a-R, a+R)$.

Hint: Show that if $x\in(a-R, a+R)$, $\lim_{n \to \infty} \frac{M_n}{n!}(x-a)^n = 0$. Why is this helpful? Functions that equal the sum of their power series are said to be 'analytic'.

Haven't worked on Taylor series in analysis class, but have this for homework. Don't quite know where to start. Any help appreciated!

Edit: Ok, so I proved the hint. Can you guide me further? I looked up the Cauchy/Langrangian forms of the remainder, but how do I link that to the "hint"? We can't say f(x) converges because of the hint can we?

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