I have the following problem.

If ($a_n$) is a sequence of real numbers such that $ \lim\limits_{n\to \infty}a_n=g$

Show that

$ \lim\limits_{x\to \infty} e^{-x}\sum\limits_{n=0}^\infty a_n \frac{x^n}{n!}=g$

I was wondering if there is an easy way to do this problem, I think I could do it using the definition of the limit of a sequence but because the power series looks so much like the power series for exponential, I feel there should be a better method.


  • $\begingroup$ For any polynomial $p(x), \lim_{x\to\infty}e^{-x}p(x)=0$ $\endgroup$
    – saulspatz
    Apr 15 '18 at 18:10

Given an $\epsilon>0$, there is some $N$ such that $|a_{n}-g|<\epsilon$ for all $n\geq N$, then for $x>0$ \begin{align*} \left|e^{-x}\sum_{n\geq N}a_{n}\dfrac{x^{n}}{n!}-e^{-x}\sum_{n\geq N}g\dfrac{x^{n}}{n!}\right|\leq e^{-x}\sum_{n\geq N}|a_{n}-g|\dfrac{x^{n}}{n!}\leq \epsilon e^{-x}\sum_{n\geq N}\dfrac{x^{n}}{n!} \end{align*}

but we know that \begin{align*} e^{x}=\sum_{n=0}^{\infty}\dfrac{x^{n}}{n!}\geq\sum_{n\geq N}\dfrac{x^{n}}{n!}>0, \end{align*} so \begin{align*} \left|e^{-x}\sum_{n\geq N}a_{n}\dfrac{x^{n}}{n!}-e^{-x}\sum_{n\geq N}g\dfrac{x^{n}}{n!}\right|\leq\epsilon. \end{align*}

Since \begin{align*} e^{-x}\sum_{n=1}^{\infty}a_{n}\dfrac{x^{n}}{n!}-g&=e^{-x}\sum_{n=1}^{N-1}(a_{n}-g)\dfrac{x^{n}}{n!}+e^{-x}\sum_{n\geq N}a_{n}\dfrac{x^{n}}{n!}-e^{-x}\sum_{n\geq N}a_{n}\dfrac{x^{n}}{n!}, \end{align*} we have \begin{align*} \left|e^{-x}\sum_{n=1}^{\infty}a_{n}\dfrac{x^{n}}{n!}-g \right|\leq e^{-x}\sum_{n=1}^{N-1}|a_{n}-g|\dfrac{x^{n}}{n!}+\epsilon \end{align*} taking $x\rightarrow\infty$, we have \begin{align*} \limsup_{x\rightarrow\infty}\left|e^{-x}\sum_{n=1}^{\infty}a_{n}\dfrac{x^{n}}{n!}-g \right|\leq\epsilon, \end{align*} since $\epsilon>0$ is arbitrary, the assertion follows.


Same ideas as the answer of user284331, but perhaps shorter.

Verify that it's enough to treat the case $g=0.$ ($g$ is weird notation for a limit btw, but fine.) Let $\epsilon>0.$ Then there exists $N$ such that $|a_n|<\epsilon$ for $n>N.$ Thus

$$\tag 1 |e^{-x}\sum_{n=0}^{\infty}a_n\frac{x^n}{n!}| \le |e^{-x}\sum_{n=0}^{N}a_n\frac{x^n}{n!}| + |e^{-x}\sum_{n>N}a_n\frac{x^n}{n!}|.$$

The first expression on the right of $(1)$ $\to 0$ because exponential decay overwhelms polynomial growth as $x\to \infty.$. The second expression on the right of $(1)$, for $x>0,$ is less than

$$\epsilon e^{-x}\sum_{n>N}\frac{x^n}{n!} \le \epsilon e^{-x}\cdot e^{x} = \epsilon.$$

It follows that the $\limsup_{x\to\infty}$ of the left side of $(1)$ is $\le \epsilon.$ Because $\epsilon$ is arbitrarily small, this $\limsup$ equals $0,$ and hence the limit of $(1)$ is $0$ as desired.


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