证明
$$\lim_{n\rightarrow\infty}{\frac{n!}{n^{n}}\left(\sum_{k=0}^{n}{\frac{n^{k}}{k!}}-\sum_{k=n+1}^{\infty}{\frac{n^{k}}{k!}}  \right)}=\frac{4}{3}$$

（西西）
我们有
$$e^{n}=\sum_{k=0}^{n}{\frac{n^{k}}{k!}}+\sum_{k=n+1}^{\infty}{\frac{n^{k}}{k!}}=\sum_{k=0}^{n}{\frac{n^{k}}{k!}}+\frac{1}{n!}\int_{0}^{n}e^{t}(n-t)^{n}dt$$
所以
$$\sum_{k=n+1}^{\infty}{\frac{n^{k}}{k!}}=\frac{1}{n!}\int_{0}^{n}e^{t}(n-t)^{n}dt$$
$$\sum_{k=0}^{n}{\frac{n^{k}}{k!}}=e^{n}-\frac{1}{n!}\int_{0}^{n}e^{t}(n-t)^{n}dt$$
因此，只要计算
$$\lim_{n\to\infty}\frac{n!}{n^{n}}\left(e^{n}-\frac{2}{n!}\int_{0}^{n}e^{t}(n-t)^{n}dt \right)$$
下面来估计
$$\int_{0}^{n}e^{t}(n-t)^{n}dt$$
我们有
$$\begin{align*}  \int_{0}^{n}e^{t}(n-t)^{n}dt&=n^{n+1}\int_{0}^{1}e^{nz}(1-z)^{n}dz\\ &=n^{n+1}\int_{0}^{1}e^{n(z+\ln(1-z))}dz\\ &=n^{n+1}\int_{0}^{1}e^{-\frac{1}{2}nz^{2}-\frac{1}{3}nz^{3}+o(nz^{3})}dz\\ &=n^{n+1}\int_{0}^{1}e^{-\frac{1}{2}nz^{2}}\left(1-\frac{1}{3}nz^{3}+o(nz^{3}) \right)dz \end{align*}$$
$$\begin{align*}  \frac{n!}{n^{n}}\left(e^{n}-\frac{2}{n!}\int_{0}^{n}e^{t}(n-t)^{n}dt \right)&= \frac{n!e^{n}}{n^{n}}-2n\left[\int_{0}^{1}e^{-\frac{1}{2}nz^{2}}\left(1-\frac{1}{3}nz^{3}+o(nz^{3}) \right)dz \right]\\ &=\left(\sqrt{2\pi{n}}e^{\frac{\theta_{n}}{12n}}-2n\int_{0}^{1}e^{-\frac{1}{2}nz^{2}}dz\right)+2n\int_{0}^{1}e^{-\frac{1}{2}nz^{2}}\left(\frac{1}{3}nz^{3}+o(nz^{3})  \right)dz \end{align*}$$
其中$\theta_{n}\in(0,1)$
显然有
$$\lim_{n\to\infty}\left(\sqrt{2\pi{n}}e^{\frac{\theta_{n}}{12n}}-2n\int_{0}^{1}e^{-\frac{1}{2}nz^{2}}dz\right)=0$$
$$\lim_{n\to\infty}2n\int_{0}^{1}e^{-\frac{1}{2}nz^{2}}\left(\frac{1}{3}nz^{3}+o(nz^{3})\right)dz=\lim_{n\to\infty}\frac{4}{3}\left(\int_{0}^{\frac{n}{2}}e^{-z}zdz+o\left( \frac{1}{n}\right)\right)=\frac{4}{3}$$
所以
$$\lim_{n\rightarrow\infty}{\frac{n!}{n^{n}}\left(\sum_{k=0}^{n}{\frac{n^{k}}{k!}}-\sum_{k=n+1}^{\infty}{\frac{n^{k}}{k!}}  \right)}=\frac{4}{3}$$