POTW Proof of Asymptotic Equality: $\sum_{n=0}^\infty a_n\frac{x^n}{n!} \sim ae^x$

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The discussion focuses on proving the asymptotic equality of the series $$\sum_{n=0}^\infty a_n\frac{x^n}{n!}$$ as it approaches $$ae^x$$ when the limit of the sequence $(a_n)$ is $a$. Participants explore the conditions under which this asymptotic behavior holds true, particularly emphasizing the significance of the convergence of $(a_n)$. The proof involves analyzing the behavior of the series and applying relevant mathematical theorems. The conclusion reinforces that the series indeed behaves asymptotically like $ae^x$ as $x$ approaches infinity. This highlights the relationship between sequences and exponential functions in asymptotic analysis.
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If ##(a_n)## is a sequence of real numbers with ##\lim a_n = a##, show that $$\sum_{n = 0}^\infty a_n\frac{x^n}{n!} \sim ae^x$$ as ##x\to \infty##.
 
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For any ##\epsilon >0## there exist ##N## such that if ##n>N##, ##|a_n-a| < \epsilon##.

|(RHS-LHS)/RHS |=
|\frac{\sum_{n=0}^\infty (a-a_n)\frac{x^n}{n!}}{ae^x}|&lt; \frac{\sum_{n=0}^N |(a-a_n)\frac{x^n} {n!}|+\epsilon \sum_{n=N+1}^\infty \frac{x^n}{n!}}{|a|e^x}
= \frac{\sum_{n=0}^N (|a-a_n|-\epsilon)\frac{x^n} {n!}+\epsilon e^x}{|a|e^x}\rightarrow \frac{\epsilon}{|a|}
##\epsilon## can be taken as small as we like. So the given asymptotic equality is proved.
 
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