Basic question about real-analytic functions

A counterexample is $$ f(x) = xe^x = \sum_{n=1}^\infty \frac{x^n}{(n-1)!} $$ where $$ n! a_n = \frac{n!}{(n-1)!} = n $$ is unbounded. Another example is $$ g(x) = e^{x^2} = \sum_{n=0}^\infty \frac{x^{2n}}{n!} $$ where for the even-indexed coefficients $$ (2n)!a_{2n} = \frac{(2n)!}{n!} = (n+1)(n+2)\ldots(2n) $$ grows faster than any polynomial.

What you can say about the growth is that $\lim_{n\to\infty} \sqrt[n]{|a_n|} = 0$ if $\sum_{n=0}^\infty a_n x^n $ is convergent for all $x \in \Bbb R$, this follows from the formula for the radius of convergence of a power series.

The function $f(x)=\dfrac{1}{1+x^2}$ is real analytic on the whole line, but the Taylor series of $f$ centered at $0$ is

$$1-x^2+x^4- x^6 +\cdots.$$

Thus $|a_n|=1$ for all even $n.$