Is there a $C^{\infty}$ function $f$ such that the radius of convergence of $\sum_{n=0}^\infty \frac{f^{(n)}(x_0)}{n!} (x-x_0)^n$ is $0$?

Question

Is there a $C^{\infty}$ function $f$ such that the radius of convergence of $\sum_{n=0}^\infty \frac{f^{(n)}(x_0)}{n!} (x-x_0)^n$ is $0$?

By the way, I know the following fact:

If $f(x) = \sum_{n=0}^\infty a_n x^n$ for $x \in (-R, R)$ for a positive real number $R$, then $f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n$ for $x \in (-R, R)$.

So, if there is a $C^{\infty}$ function $f$ such that the radius of convergence of $\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n$ is $0$, then we cannot write $f(x) = \sum_{n=0}^\infty a_n x^n$ for $x \in (-R, R)$ for a positive real number $R$.

Answer 1

Yes. Given any sequence $(a_n)_{n\in\mathbb Z^+}$ of real numbers, there is a $C^\infty$ function $f$ such that it's Taylor series centered at $0$ is $\sum_{n=0}^\infty a_nx^n$. So, take, say, $a_n=n!$ and you're done.

The statement that I mentioned is known as Borel's theorem on power series and you will find a proof at, say, Ádám Besenyei's article Peano's unnoticed proof of Borel's theorem.

Answer 2

Choose any sequence a_n such that $$ \mathop {\lim }\limits_{n \to \infty } a_n = 0 $$ According to Du-Boys Reymond the function

$$ f(x) = \sum\limits_{n = 1}^\infty {\left( { - 1} \right)^{n + 1} } \frac{{x^{2n} }} {{\left[ {\left( {2n} \right)!\left( {x^2 + a^2 _n } \right)} \right]}} $$

is $$ C^\infty $$ but its Taylor series in x=0 has convergence radius r=0.