$$\lim \limits_{n\mathop\to\infty}\frac{1}{e^n}\sum \limits_{k\mathop=0}^n\frac{n^k}{k!} $$
I thought this limit was obviously $1$ at first but approximations on Mathematica tells me it's $1/2$. Why is this?
Answer
In this answer, it is shown that
$$
\begin{align}
e^{-n}\sum_{k=0}^n\frac{n^k}{k!}
&=\frac{1}{n!}\int_n^\infty e^{-t}\,t^n\,\mathrm{d}t\\
&=\frac12+\frac{2/3}{\sqrt{2\pi n}}+O(n^{-1})
\end{align}
$$
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