calculus - Closed Form for $~limlimits_{ntoinfty}~sqrt ncdot(-e)^{-n}cdotsumlimits_{k=0}^nfrac{(-n)^k}{k!}$

$\qquad\qquad\qquad$ Does $~\displaystyle\lim_{n\to\infty}\frac{\sqrt n}{(-e)^n}\cdot\sum_{k=0}^n\frac{(-n)^k}{k!}~$ possess a closed form expression ?

Inspired by this frequently asked question, I wondered what would happen if the sum were allowed to alternate. Numerically, it seems to converge to a value around $~\dfrac15$ . Unfortunately, I wasn't truly able to grasp any of the various approaches used to evaluate the other related limit $($yes, I actually read carefully through all of them$)$, so I haven't been successful in developing a viable method for expressing this one either. $($Perhaps a new, insightful answer will also help me cast some fresh light on older ones ?$)$

Answer

It was shown in the answers to this question that

$$e^{-x}\sum_{k=0}^n\frac{x^k}{k!} = \frac{1}{n!}\int_x^\infty e^{-t}\,t^n\,dt,$$

so setting $x=-n$ we have

$$\begin{align} e^{n}\sum_{k=0}^n\frac{(-n)^k}{k!} &= \frac{1}{n!}\int_0^\infty e^{-t}\,t^n\,dt + \frac{1}{n!}\int_{-n}^0 e^{-t}\,t^n\,dt \\ &= 1 + \frac{(-1)^n}{n!} \int_0^n e^u u^n\,du \\ &= 1 + \frac{(-1)^n n^{n+1}}{n!} \int_0^1 e^{n [v+\log v]}\,dv. \tag{$*$} \end{align}$$

The quantity $v+\log v$ is increasing and so has a maximum at $v=1$, and near there

$$v+\log v = 1 + 2(v-1) + O\!\left((v-1)^2\right).$$

By the Laplace method we therefore have

$$\int_0^1 e^{n [v+\log v]}\,dv \sim \int_{-\infty}^1 e^{n[1 + 2(v-1)]}\,dv = \frac{e^n}{2n}.$$

Using this and Stirling's formula

$$n! \sim \left(\frac{n}{e}\right)^n \sqrt{2\pi n}$$

we deduce from $(*)$ that

$$\sum_{k=0}^n\frac{(-n)^k}{k!} \sim \frac{(-e)^n}{2\sqrt{2\pi n}}.$$

The limit in the question is

$$\lim_{n\to\infty}\frac{\sqrt n}{(-e)^n}\cdot\sum_{k=0}^n\frac{(-n)^k}{k!} = \frac{1}{2\sqrt{2\pi}} = 0.199471\ldots$$