Roger Smith

2022-01-17

Calculate: $\underset{n\to \infty }{\mathrm{lim}}{e}^{-n}\sum _{k=0}^{n}\frac{{n}^{k}}{k!}$

Terry Ray

$\underset{n\to \mathrm{\infty }}{lim}\left[{e}^{-n}\sum _{k=0}^{n}\frac{{n}^{k}}{k!}\right]$
$=\underset{n\to \mathrm{\infty }}{lim}\left[{e}^{-n}\left[\sum _{k=0}^{n}\mathrm{exp}\left(k\mathrm{ln}\left(n\right)-\mathrm{ln}\left(k!\right)\right)\right]$
$=\underset{n\to \mathrm{\infty }}{lim}\left[{e}^{-n}\sum _{k=0}^{n}\mathrm{exp}\left(n\mathrm{ln}\left(n\right)-\mathrm{ln}\left(n!\right)-\frac{1}{2n}{\left[k-n\right]}^{2}\right)\right]$
$=\underset{n\to \mathrm{\infty }}{lim}\left[{e}^{-n}\frac{{n}^{n}}{n!}\sum _{k=0}^{n}\mathrm{exp}\left(-\frac{1}{2n}{\left[k-n\right]}^{2}\right)\right]$
$=\frac{1}{2}\underset{n\to \mathrm{\infty }}{lim}\left[\frac{\sqrt{2\pi }{n}^{n+\frac{1}{2}}{e}^{-n}}{n!}\right]=\frac{1}{2}$

Lakisha Archer

The sum is related to the partial exponential sum, and thus to the incomplete gamma function,
${e}^{-n}\sum _{k=0}^{n}\frac{{n}^{k}}{k!}={e}^{-n}{e}^{n}\left(n\right)$
$=\frac{\mathrm{\Gamma }\left(n+1,n\right)}{\mathrm{\Gamma }\left(n+1\right)}$
since ${e}_{n}\left(x\right)=\sum _{k=0}\frac{{x}^{k}}{k}\ne {e}^{x}\mathrm{\Gamma }\frac{n+1,x}{\mathrm{\Gamma }}\left(n+1\right)$. But
$\mathrm{\Gamma }\left(n+1,n\right)=\sqrt{2\pi }{n}^{n+\frac{1}{2}}{e}^{-n}\left(\frac{1}{2}+\frac{1}{3}\sqrt{\frac{2}{n\pi }}+O\left(\frac{1}{n}\right)\right)$
The first term in the asymptotic expansion for $\mathrm{\Gamma }\left(n+1,n\right)={\int }_{n}^{\mathrm{\infty }}dt{t}^{n}{e}^{-t}$
The higher order terms are in principle straightforward to compute. Using Stirlings

alenahelenash

On this page there is a nice collection of evidence. I add another proof which also uses the Stirling formula. ${e}^{-n}\sum _{k=0}^{n}\frac{{n}^{k}}{k!}={e}^{-n}\sum _{k=0}^{n}\frac{{k}^{k}\left(n-k{\right)}^{n-k}}{k!\left(n-k\right)!}$ $\underset{n\to \mathrm{\infty }}{lim}{e}^{-n}\sum _{k=1}^{n-1}\frac{{e}^{k}{e}^{n-k}}{\sqrt{2\pi k\left(1+O\left(1/k\right)\right)}\sqrt{2\pi \left(n-k\right)\left(1+O\left(1/\left(n-k\right)\right)\right)}}$ $\underset{n\to \mathrm{\infty }}{lim}\frac{1}{2\pi }\frac{1}{n}\sum _{k=1}^{n-1}\frac{1}{\sqrt{\frac{k}{n}\left(1-\frac{k}{n}\right)}}=\frac{1}{2\pi }{\int }_{0}^{1}\frac{dx}{\sqrt{x\left(1-x\right)}}=\frac{\mathrm{\Gamma }\left(\frac{1}{2}{\right)}^{2}}{2\pi \mathrm{\Gamma }\left(1\right)}=\frac{1}{2}$