Assume that $f:(a,b) \rightarrow \mathbf{R}$, $x_0 \in (a,b)$ and $f(x)=\sum_{n=0}^\infty a_n (x-x_0)^n$ for $x \in (a,b)$ and radius of convergence $R$ of this power series is infinite. (Then $f$ is smooth, $a_n=\frac{f^{(n)}(x_0)}{n!}$ for $n=0, 1, 2, \ldots$ and $\frac{1}{R}=\lim_{n \rightarrow \infty} \sqrt[n] { \frac{ |f^{(n)} (x_0)|} {n!} }=0$). Is it true that
$ \lim_{n \rightarrow \infty} \sqrt[n] { \frac{\sup_{x \in K} |f^{(n)} (x)|} {n!} }=0$ for every compact $K \subset (a,b)$ ?
Thanks.