A question (Are derivatives actually bounded?) has been asked on Stackexchange as to whether there exists $f\in C^{\infty}$ such that $f^{(n)}(0)=(n!)^2$. Obviously $f$ is not analytic but the respondent cites Borel's theorem as implying an affirmative answer.
For fun, I decided to construct one such $f$, and I hit on the following:
$g_n(x)=\exp\frac{\exp\frac{1}{n|x|-1}}{-|x|}$ for $0<|x|<\frac{1}{n}$
$f_n(x)=x^nn![1-g_n(x)]$ for $0<|x|<\frac{1}{n}$; anywhere else, $f_n(x)=0$
$f(x)=\sum_{n=1}^{\infty} f_n(x)$
Are there functions with that property which are easier to verify? Because it took me ages to produce what I believe to be a proof that the k-th derivative of $\sum_{n=1}^{\infty}x^nn!g_n(x)$ is a uniformly convergent series.
And is $f$ the required function?
