Is there a $C^\infty$ function $f$ s. t. the convergent radius of Maclaurin series of $f$ is $0$?

Question

To be precisely, let $f$ be a $C^\infty$ function defined on $(-\epsilon,\epsilon)$, where $\epsilon>0$.

Question: Can the power series $\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n$ have convergent radius $R=0$?

I think there might be some example satisfies $R=0$, but I cannot construct it, since it must be some non-elementary function and cannot be writen down in simple formula.

Any comments and ideas are welcome.

Answer 1

There is a theorem by Borel that asserts that every formal power series is the MacLaurin series of some function that is $C^{\infty}$ all over $\mathbb{R}$, i.e. the map $C^{\infty}(\mathbb{R}) \to \mathbb{R}[[x]]$ defined by $$ f \mapsto \sum_{n \geq 0} \frac{f^{(n)}(0)}{n!}x^n $$ is surjective. See here.