In this article on Non-analytic smooth function seems to offer the following argument:
$|F^{(n)}(x_o)| \geq e^{-\sqrt{n}}n^n + O(q^n) \ $ implies $ \ \displaystyle \limsup_{n \rightarrow \infty} \left( \frac{|F^{(n)}(x_o)|}{n!} \right)^{1/n} = \infty$
where $q = 2^m$ for some positive integer $m$. My initial thought was the implication followed by comparison since $$ \limsup_{n \rightarrow \infty} \left(\frac{e^{-\sqrt{n}}n^n + O(q^n)}{n!}\right)^{1/n} = \infty \qquad (\text{false})$$ But, I believe the limit above actually converges. In particular, this example calculation in wolfram alpha and similar calculations show the $O(q^n)$-type term is irrelevant and the limit of $\left(e^{-\sqrt{n}}n^n/n! \right)^{1/n} \rightarrow e$ as $n \rightarrow \infty$. Since the limsup matches the limit when it exists we cannot conclude the limsup of $\left(\frac{e^{-\sqrt{n}}n^n + O(q^n)}{n!}\right)^{1/n}$ diverges. Therefore, there is no comparison to be made and I don't see how we are able to say anything about $\limsup_{n \rightarrow \infty} \left( \frac{|F^(n)(x_o)|}{n!} \right)^{1/n}$.
Question: is Wikipedia wrong here? Show me the error of my ways please.
That's a mistake in the wikipedia article. The given estimate using the term for $k = n$ only shows $$\limsup_{n \to \infty} \:\biggl(\frac{\lvert F^{(n)}(x_0)\rvert}{n!}\biggr)^{\frac{1}{n}} \geqslant e\,.$$
To fix it, replace $\cos (kx)$ with $\cos (k^rx)$ for your favourite $r > 1$ in the definition of $F$.
However, the function $F$ as defined in the wikipedia article is indeed nowhere analytic, as one can see using the term for $k = n^2$ for the estimate. Then we obtain
$$F^{(n)}(x_0) \geqslant e^{-n}n^{2n} - C\cdot q^n$$
and that yields
$$\biggl(\frac{\lvert F^{(n)}(x_0)\rvert}{n!}\biggr)^{\frac{1}{n}} \geqslant n\cdot \bigl(1 + o(1)\bigr)\,.$$