On Terry Tao's blog I found the following problem: https://terrytao.wordpress.com/2016/10/18/a-problem-involving-power-series/#comment-513808 I think I have a solution but I dont know how to write math on his blog so I posted my solution here.
Problem: Let $\{ a_n\}_{n\in \mathbb{N}}\subseteq \mathbb{R}$ be a bounded sequence. Suppose that the power series: $$f(x)=\overset{\infty}{\underset{n=0}{\sum}}\dfrac{a_n}{n!}x^n.$$ (which has an infinite radius of convergence) decays like $O(e^{-x})$ as $x\to\infty$ ($\underset{x\to \infty}{\lim}\dfrac{f(x)}{e^{-x}}=C$). Must the sequence ${a_n}$ be of the form ${a_n = C (-1)^n}$ for some constant ${C}$?
Here is my purely real variable solution. We have that $e^{x}$ and $f(x)$ are both analytic, so their product $\phi(x)=e^{x}\cdot f(x)$ is, therefore $\phi(x)=\overset{\infty}{\underset{n=0}{\sum}}c_n x^n$, where $c_n= a_0\cdot\frac{1}{(n-1)!}+...+\frac{a_i}{i!}\cdot\frac{1}{(n-i)!}+...+\frac{a_n}{n!}$. Since $\underset{x\to \infty}{\lim}\phi(x)=\underset{x\to \infty}{\lim}\dfrac{f(x)}{e^{-x}}=C$, for every $\epsilon>0$ there are $R>0$ and $N\in\mathbb{N}$ such that: $$\Bigg|\overset{k}{\underset{n=0}{\sum}}c_n x^n - C \Bigg|\leq \Bigg|\overset{k}{\underset{n=0}{\sum}}c_n x^n - \overset{\infty}{\underset{n=0}{\sum}}c_n x^n \Bigg|+ \Bigg|\overset{\infty}{\underset{n=0}{\sum}}c_n x^n - C \Bigg|\leq \dfrac{\epsilon}{2}+\dfrac{\epsilon}{2}=\epsilon.$$
for all $k> N$ and all $x> R$. Thus,
$$C- \epsilon < \overset{k}{\underset{n=0}{\sum}}c_n x^n <\epsilon + C, \text{ for }k> N, x > R$$ this implies that,
$$0=\underset{x\to \infty}{\lim}\dfrac{C- \epsilon}{x^k} < c_k <\underset{x\to \infty}{\lim}\dfrac{C+ \epsilon}{x^k}=0, ~\forall~ k> N.$$ From this we deduce that $\phi(x)=\overset{N}{\underset{n=0}{\sum}}c_n x^n$ and because $\underset{x\to \infty}{\lim}\phi(x)=C$, $\phi(x)$ must be constant $C.$ Hence, $f(x)=Ce^{-x}$, where the equations $a_n=C(-1)^n$ follows.
Edit: So if $f(x)$ decays like $O(e^{-x})$, means that $\phi(x)=e^{x}f(x)$ remains bounded as $x\to \infty$, let $\alpha :=\underset{x\to \infty}{\liminf}\phi(x)$, $\beta :=\underset{x\to \infty}{\limsup}\phi(x)$, and both are finite. Then $\forall\epsilon>0$, there are $R_1, R_2>0$ such that
$$\alpha - \epsilon < \overset{\infty}{\underset{n=0}{\sum}}c_n x^n <\beta + \epsilon, \text{ for all }, x > max\{R_1, R_2\}$$
Now we can take $N\in\mathbb{N}$ such that
$$\alpha - \epsilon < \overset{k}{\underset{n=0}{\sum}}c_n x^n <\beta + \epsilon, \text{ for } k>N, \text{ and } x > max\{R_1, R_2\}$$ this implies that,
$$0=\underset{x\to \infty}{\lim}\dfrac{\alpha- \epsilon}{x^k} < c_k <\underset{x\to \infty}{\lim}\dfrac{\beta+ \epsilon}{x^k}=0, ~\forall~ k> N.$$
From this we deduce that $\phi(x)=\overset{N}{\underset{n=0}{\sum}}c_n x^n$. The same argument applied to the polynomial $\phi(x)$ shows that $c_n=0$ $\forall n>0$ and $\alpha = c_0 = \beta$, therefore $\phi(x)$ must be constant $C$, for $C=c_0$ Hence, $f(x)=Ce^{-x}$, and the equations $a_n=C(-1)^n$ follows.