Closed Form for $~\lim\limits_{n\to\infty}~\sqrt n\cdot(-e)^{-n}\cdot\sum\limits_{k=0}^n\frac{(-n)^k}{k!}$ 
$\qquad\qquad\qquad$ 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 ?$)$
 A: 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
$$
\lim_{n\to\infty}\frac{\sqrt n}{(-e)^n}\cdot\sum_{k=0}^n\frac{(-n)^k}{k!} = \frac{1}{2\sqrt{2\pi}} = 0.199471\ldots
$$
