2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $n>N$.

$\endgroup$
0
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.