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

[Euge]

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 $x\to \infty$.

 

[anuttarasammyak]

Spoiler

For any $\epsilon >0$ there exist $N$ such that if $n>N$, $|a_n-a| < \epsilon$.

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

 

[mxwlljack]

If (a_n\to a), write
[
\sum_{n\ge0}a_n\frac{x^n}{n!}
=ae^x+\sum_{n\ge0}(a_n-a)\frac{x^n}{n!}.
]
Since (a_n-a\t, the second sum is (o(e^x)). Hence
[
\boxed{\sum_{n\ge0}a_n\frac{x^n}{n!}\sim ae^x\quad(x\to\infty).}
]

 

[Gavran]

If (a_n\to a), write
[
\sum_{n\ge0}a_n\frac{x^n}{n!}
=ae^x+\sum_{n\ge0}(a_n-a)\frac{x^n}{n!}.
]
Since (a_n-a\t, the second sum is (o(e^x)). Hence
[
\boxed{\sum_{n\ge0}a_n\frac{x^n}{n!}\sim ae^x\quad(x\to\infty).}
]

It seems to me that you did not show that $\sum_{n\ge0}(a_n-a)\frac{x^n}{n!}\sim0$, which is crucial to the proof.