Find the limit of this function as $x\to\infty$ Prove that if a sequence of real numbers $(a_n)$ converges to $g$, then $\text{lim}_{x\to\infty}e^{-x}\sum_{n=0}^{\infty}a_n\frac{x^n}{n!}=g$.
I'm not exactly sure how to do this. I tried looking at the partial sums firs $\sum_{n=0}^ka_n\frac{x^n}{n!}$. For a fixed $x$, I'm guessing that this converges to a function $g\cdot l(x)$ where $\frac{l(x)}{e^x}\to 0$ as $x\to\infty$. I'm not sure how to show this though. I tried to bound $|(a_1x+a_2x^2/2+...+a_kx^k/k!)-kx^kg|$, but wasn't getting anywhere. Does anyone have advice on how I should get started?
 A: Take an arbitrary $\epsilon>0.$ Then there exists an $n_0$ such that $|a_n - g|<\epsilon$ for all $n>n_0$.
$$\sum_{n=1}^{n_0} a_n \frac{x^n}{n!}+\sum_{n=n_0+1}^\infty (g-\epsilon)\frac{x^n}{n!}<\sum_{n=1}^\infty a_n \frac{x^n}{n!} < \sum_{n=1}^{n_0} a_n \frac{x^n}{n!}+\sum_{n=n_0+1}^\infty (g+\epsilon)\frac{x^n}{n!}$$
We can see that $\sum_{n=n_0+1}^\infty (g+\epsilon)\frac{x^n}{n!}=ge^x+\epsilon e^x - (g+\epsilon)\sum_{n=1}^{n_0}\frac{x^n}{n!}$. Let us denote $\sum_{n=1}^{n_0}\frac{x^n}{n!}$ as $M$ to save time. We can similarly see that $\sum_{n=1}^{n_0} a_n \frac{x^n}{n!}+\sum_{n=n_0+1}^\infty (g-\epsilon)\frac{x^n}{n!}$ is equal to $g e^x - \epsilon e^x + (g-\epsilon)M$.
So , we can write, $\displaystyle \lim_{x \to \infty}[e^{-x}(ge^x - \epsilon e^x + (g-\epsilon)M + \sum_{n=1}^{n_0}a_n \frac{x^n}{n!})]<\lim_{x \to \infty}e^{-x}\sum_{n=1}^\infty a_n \frac{x^n}{n!}<\lim_{x \to \infty}[e^{-x}(ge^x + \epsilon e^x + (g+\epsilon)M +\sum_{n=1}^{n_0}a_n \frac{x^n}{n!})]$
Clearly then, $$g-\epsilon<\lim_{x \to \infty}e^{-x}\sum_{n=1}^\infty a_n \frac{x^n}{n!}<g+\epsilon$$.
This proves the result.
