Weird limit $\lim \limits_{n\mathop\to\infty}\frac{1}{e^n}\sum \limits_{k\mathop=0}^n\frac{n^k}{k!} $ [duplicate]

$$\lim \limits_{n\mathop\to\infty}\frac{1}{e^n}\sum \limits_{k\mathop=0}^n\frac{n^k}{k!} $$

I thought this limit was obviously $1$ at first but approximations on Mathematica tells me it's $1/2$. Why is this?

Solution 1:

You made a mistake since the summation goes to $n$ and not to infinity; then $$\sum \limits_{k\mathop=0}^n\frac{n^k}{k!} \neq e^n$$ In fact $$\sum \limits_{k\mathop=0}^n\frac{n^k}{k!}=\frac{e^n \Gamma (n+1,n)}{\Gamma (n+1)}$$ and so the limit you look for is then the limit of $\frac{\Gamma (n+1,n)}{ \Gamma (n+1)}$ when $n$ goes to infinity and this limit is $\frac{1}{2}$ as Mathematica told you.

Solution 2:

In this answer, it is shown that $$ \begin{align} e^{-n}\sum_{k=0}^n\frac{n^k}{k!} &=\frac{1}{n!}\int_n^\infty e^{-t}\,t^n\,\mathrm{d}t\\ &=\frac12+\frac{2/3}{\sqrt{2\pi n}}+O(n^{-1}) \end{align} $$