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