Let $f:\mathbb{R}\to\mathbb{R}$ to be a real-analytic function in the whole line. Then, we can write $f(x)$ as the (convergente) power series $$ f(x)=\sum_{n=0}^\infty a_nx^n, \quad \hbox{where}\quad a_n:=\dfrac{f^{(n)}(0)}{n!}, $$ where $f^{(n)}(0)$ denotes its $n$-th derivative evaluated at $x=0$. In a previous post I have asked if the sequence of coefficients $a_n$ always decays as $\sim\tfrac{1}{n!}$ when $n$ goes to $\infty$. After two great and basic counter-examples of another member I realize that this (too) strong property does not hold in general. However, as you can see below, both counterexamples still exhibits exponential decay! So I am wondering if the following weaker property holds: For any $m\in\mathbb{N}$, there exists a constant $C>0$ such that for all $n=1,2,3,...$ we have $$ \vert a_n\vert \leq \dfrac{C}{n^m} \ ? $$ Of course, this is a weaker statement than the one of my previous post (decay of order $\tfrac{1}{n!}$), and also weaker than exponential decay.
Previous counter-examples: For the sake of completeness let me copy-paste both previous counter-examples:
1.- $$ f(x) = xe^x = \sum_{n=1}^\infty \frac{x^n}{(n-1)!} \quad \implies a_n = \frac{1}{(n-1)!} $$ 2.- $$ g(x) = e^{x^2} = \sum_{n=0}^\infty \frac{x^{2n}}{n!} \quad \implies \quad a_{2n} = \frac{1}{n!}. $$
Last comment: Notice that in both cases we have $$ \lim_{n\to+\infty} a_n\exp(n)=0. $$