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

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

• I had a typo in the original post, where in the first display $a_n$ should have been $n$ - fixed now – Richard Aug 31 '18 at 15:32
• Is $a_n > 0$ for all $n$? – Daniel Fischer Aug 31 '18 at 15:36
• @DanielFischer Yes – Richard Aug 31 '18 at 15:39
• Perhaps it was one of these questions that you saw. – Antonio Vargas Aug 31 '18 at 18:21
• Why does this not tend to 1? Do you have the original proof? – Henry Lee Aug 31 '18 at 18:53

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 $$\sum_{k = 0}^{n} \frac{a_n^k}{k!} \sim e^{a_n}\,.$$
• How is the step $e^{a_n} - \sum_{k=0}^{n} \frac{a_n^k}{k!}= \sum_{k=n+1}^{\infty} \frac{a_n^k}{k!}$ motivated taking into account that $a_n$ is varying? – user Aug 31 '18 at 15:54
• @gimusi It depends on $n$, but not on $k$. – Daniel Fischer Aug 31 '18 at 15:56
• But isn't it equivalent to assume $e^{a_n} = \sum_{k = 0}^{\infty} \frac{a_n^k}{k!}$? – user Aug 31 '18 at 16:00
• But that's a given. $$e^x = \sum_{k = 0}^{\infty} \frac{x^k}{k!}$$ is the definition of $e^x$. (Yes, it is possible to define it differently, then one needs to prove the power series expansion, but that's standard.) – Daniel Fischer Aug 31 '18 at 16:03
• Were you perhaps thinking about $$\sum_{k = 0}^{n} \frac{a_k^k}{k!}\,,$$ @gimusi? – Daniel Fischer Aug 31 '18 at 16:06