Basic question about real-analytic functions

Question

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 $x=0$. I am wondering if the following property is true: There exists a constant $C>0$ such that for all $n=0,1,2,...$ we have $$ \vert a_n\vert \leq \dfrac{C}{n!} \ ? $$ At first glance I was thinking that this should be true, but then I kept thinking that I have no control on the growth of $f^{(n)}(0)$ as $n$ growths. However, since the series is convergent, $a_n$ should always goes to zero, so... but that doesn't mean that the decay-rate must to be of order $\tfrac{1}{n!}$, right? That is actually what motivated my question, the decay at rate $\tfrac{1}{n!}$. In other words (if you want), can exist a real-analytic function so that the sequence of $f^{n}(0)$ behaves as (for example) $$ f^{(n)}(0)=\dfrac{n!}{n^{10}} \quad \hbox{so that} \quad a_n=\dfrac{1}{n^{10}}? $$ This should also produce a convergent series, at least for $x\in (-1,1)$ small right?

Answer 1

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.

Answer 2

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 $n.$