How to prove that if $a_n=o(n) $ then $\sum_{k=0}^n\frac{a_n^k}{k!}\sim e^{a_n} $

Question

I recall a question where it was proved that $$\lim_{n\to\infty}e^{-n}\sum_{k=0}^n\frac{n^k}{k!}=\frac12.$$This seems to remain true when $n$ is replaced by $n+c $ where $c$ is some constant, while apparently, if $a_n=o(n)$, $$\sum_{k=0}^n\frac{a_n^k}{k!}\sim e^{a_n}.$$How to prove it? Maybe the method used in the related question is also useful here, but I couldn't find it

Answer 1

For every $0 < \varepsilon < e^{-1}$, choose $N(\varepsilon)$ such that $a_n < \varepsilon n$ for $n \geqslant N(\varepsilon)$. Then we have \begin{align} e^{a_n} - \sum_{k = 0}^{n} \frac{a_n^k}{k!} &= \sum_{k = n+1}^{\infty} \frac{a_n^k}{k!} \\ &< \sum_{k = n+1}^{\infty} \frac{(\varepsilon n)^k}{k!} \\ &< \varepsilon^{n+1}\sum_{k = n+1}^{\infty} \frac{n^k}{k!} \\ &< \varepsilon^{n+1} e^n \\ &< \varepsilon \end{align} for $n \geqslant N(\varepsilon)$. So we even have $$e^{a_n} - \sum_{k = 0}^{n} \frac{a_n^k}{k!} \to 0\,,$$ not only $$\sum_{k = 0}^{n} \frac{a_n^k}{k!} \sim e^{a_n}\,.$$