Show that there is an infinitely differentiable function f(x), such that the radius of convergence of its Taylor series equals to 0

Question

This is an exercise from my introduction to real analysis course, this does not count for marks. I saw several people said $\sum\limits_{n = 0}^{\infty}n!x^{n}$, I mean this power series clearly has a radius of convergence that is 0, but how can one know that it is some function's Taylor series? Or is there any other example? Please advise, thank you.

Answer 1

I don't know any explicit example but it is well known that there is a $C^{\infty}$ function $f$ with $f^{(n)}(0)=(n!)^{2}$ for all $n$. [ We can in fact prescribe the derivatives arbitrarily]. Obviously we cannot define $f$ by the series $\sum (n!)x^{n}$. The existence of $f$ is special case of Borel's Lemma.