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?
Take an arbitrary $\epsilon&gt;0.$ Then there exists an $n_0$ such that $|a_n - g|&lt;\epsilon$ for all $n&gt;n_0$.

$$\sum_{n=1}^{n_0} a_n \frac{x^n}{n!}+\sum_{n=n_0+1}^\infty (g-\epsilon)\frac{x^n}{n!}&lt;\sum_{n=1}^\infty a_n \frac{x^n}{n!} &lt; \sum_{n=1}^{n_0} a_n \frac{x^n}{n!}+\sum_{n=n_0+1}^\infty (g+\epsilon)\frac{x^n}{n!}$$

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, $\displaystyle \lim_{x \to \infty}[e^{-x}(ge^x - \epsilon e^x + (g-\epsilon)M + \sum_{n=1}^{n_0}a_n \frac{x^n}{n!})]&lt;\lim_{x \to \infty}e^{-x}\sum_{n=1}^\infty a_n \frac{x^n}{n!}&lt;\lim_{x \to \infty}[e^{-x}(ge^x + \epsilon e^x + (g+\epsilon)M +\sum_{n=1}^{n_0}a_n \frac{x^n}{n!})]$

Clearly then, $$g-\epsilon&lt;\lim_{x \to \infty}e^{-x}\sum_{n=1}^\infty a_n \frac{x^n}{n!}&lt;g+\epsilon$$.

This proves the result.
$$e^{-x}\sum_{n=0}^{\infty}a_n\frac{x^n}{n!}=e^{-x}\sum_{n=0}^{\infty}(a_n-g+g)\frac{x^n}{n!}=e^{-x}\left(\sum_{n=0}^{\infty}b_n\frac{x^n}{n!}\right)+g$$

where $b_n:=a_n-g$ tends to $0$, and we need to show that the first term tends to zero.

Then, for any $\epsilon&gt;0$, there is an $N$ such that $n\ge N\implies|b_n|&lt;\epsilon$ and

$$\left|\sum_{n=0}^{\infty}b_n\frac{x^n}{n!}\right|&lt;\left|\sum_{n=0}^{N-1}b_n\frac{x^n}{n!}\right|+\epsilon\sum_{n=N}^{\infty}\frac{x^n}{n!}&lt;|S_N|+\epsilon e^x.$$

So for any $\epsilon$, there is a finite $S_N$ such that

$$\lim_{x\to\infty}e^{-x}\left|\sum_{n=0}^{\infty}b_n\frac{x^n}{n!}\right|&lt;\lim_{x\to\infty}e^{-x}(|S_N|+\epsilon e^x)=\epsilon.$$