This question already has an answer here:

If the power series $\sum_{n\in\mathbb N}a_n\frac{x^n}{n!}$ converges for all real $x$, then the answer is trivially yes.

If the power series above has radius of convergence $3r>0$, let $\psi_r\in C^\infty(\mathbb R)$ be supported in $[-r,r]$ with $\psi_r(0)=1$ and $\psi_r^{(n)}(0)=0$ for $n>0$, and observe that $$ h(x):=\begin{cases} \psi_r(x)\sum_{n\in\mathbb N}a_n\frac{x^n}{n!}&\text{if }x\in [-2r,2r],\\ 0&\text{otherwise} \end{cases} $$ does the job. (For the sake of completeness, I construct such a $\psi_r$ below.)

The case where $\sum_{n\in\mathbb N}a_n\frac{x^n}{n!}$ converges only at $x=0$ (e.g., if $\{a_n\}$ is something awful like $\{(n!)^2\}$) is the one I'm stuck on.

Here we construct the function $\psi_r$ used in the second case above: Define $$\begin{align} f(x)&=\begin{cases} e^{\frac {-1}{x}}&\text{if }x>0,\\ 0&\text{if }x\leq 0 \end{cases}\\ g(x)&=\frac {f(x)}{f(x)+f(1-x)} \end{align}$$ as in this wikipedia article. Then $$\begin{align} \psi(x)&:=1-g(x)-g(-x) \end{align}$$ is a $C^\infty$ function with support $[-1,1]$ satisfying $\psi(0)=1$ and $\psi^{(n)}(0)=0$ for all $n>0$ (this much is clear from the list of properties of $g$ given in the linked article). Take $\psi_r(x)=\psi(x/r)$, and observe that $\psi_r$ does the job.


marked as duplicate by Julián Aguirre real-analysis Jul 17 '18 at 22:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Browse other questions tagged or ask your own question.