# Computing the limit of function containing a power series.

Prove that if the sequence $a_{n}$ of real numbers converges to a finite limit; \begin{align} \lim_{n \rightarrow \infty} a_{n} = g, \end{align} then \begin{align} \lim_{x \to \infty} \left({\rm e}^{-x}\sum_{n = 0}^{\infty}a_{n}\,{x^{n} \over n!}\right) = g. \end{align} The initial observation is the power series of $e^{x}$ is given by \begin{align} e^{x} = \sum_{n = 0}^{\infty} \frac{x^{n}}{n!}. \end{align} I want to use summation by parts somehow while using some sort of telescoping technique. Is this the right technique? How do I get started with this?

Note that for $x>0$ you have $$\lvert e^{-x}\sum_{n=0}^\infty a_n\frac{x^n}{n!}-g\rvert\leq e^{-x}\sum_{n=0}^\infty \lvert a_n-g\rvert\frac{x^n}{n!}= e^{-x}\sum_{n=0}^N \lvert a_n-g\rvert\frac{x^n}{n!}+e^{-x}\sum_{n>N}\lvert a_n-g\rvert\frac{x^n}{n!}<Me^{-x}\sum_{n=0}^N \frac{x^n}{n!}+\epsilon e^{-x}\sum_{n>N}\frac{x^n}{n!}=Me^{-x}\sum_{n=0}^N \frac{x^n}{n!}+\epsilon e^{-x}(e^x-\sum_{n=0}^N \frac{x^n}{n!})=(M-\epsilon)e^{-x}\sum_{n=0}^N \frac{x^n}{n!}+\epsilon \to\epsilon$$ as $x\to +\infty$, where $M=\sup_n\lvert a_n-g\rvert$, $\epsilon>0$ is small as we would and $N=N(\epsilon)$ is such that $\lvert a_n-g\rvert<\epsilon$ for $n>N$.

For each positive integer $N$, \begin{align} \limsup_{x\rightarrow\infty}\left|e^{-x}\left(\sum_{n=0}^{\infty}a_{n}\frac{x^{n}}{n!}-g\right)\right| & = \limsup_{x\rightarrow\infty}\left|e^{-x}\left(\sum_{n=0}^{\infty}a_{n}\frac{x^{n}}{n!}-\sum_{n=0}^{\infty}g\frac{x^{n}}{n!}\right)\right| \\ & = \limsup_{x\rightarrow\infty}\left|e^{-x}\sum_{n=1}^{\infty}(a_{n}-g)\frac{x^{n}}{n!}\right| \\ & = \limsup_{x\rightarrow\infty}\left|e^{-x}\sum_{n=N}^{\infty}(a_{n}-g)\frac{x^{n}}{n!}\right| \\ & \le \limsup_{x\rightarrow\infty}e^{-x}(\sup_{n\ge N}|a_{n}-g|)\sum_{n=N}^{\infty}\frac{x^{n}}{n!} \\ & \le \limsup_{x\rightarrow\infty}\left( \sup_{n\ge N}|a_{n}-g|\right) \\ & = \sup_{n\ge N}|a_{n}-g| \end{align} Because this holds for every $N$, then the result follows.

• Thanks! How can you just switch from n = 1 to n = N without changing anything? – DRich Jan 12 '14 at 2:11
• @DRich: Because the $\lim_{x\rightarrow\infty}$ of any fixed finite sum is 0 because of the $e^{-x}$ term. – DisintegratingByParts Jan 12 '14 at 3:13