# 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 ?$)$

• i'm wondering if one can make such nice probalistic arguments as in the linked question. It seems a little bit tricky because we can't just apply the properties of poisson processes (basicially because $n$ is negative inside the sum). But i'm far from an expert in this field so who knows.. Oct 4 '15 at 12:19

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

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

• nice (+1), i was working on the same answer but you beat me for approximatly five minutes! Gratulations! :) Oct 4 '15 at 11:44
• Im pretty impressed. I would like to learn more about this Laplace method to solve these problems :) Do you know any good resources? Oct 4 '15 at 11:45
• @CalvinKhor I learned it from de Bruijn's Asymptotic Methods in Analysis and Miller's Applied Asymptotic Analysis, both of which are very rigorous and informative. Oct 4 '15 at 11:50
• @Lucian I should mention that the poor approximation of $v+\log v$ by $2v-1$ away from $v=1$ doesn't matter when it comes to the limiting behavior. For large $n$ the main contribution to the integral comes from an arbitrarily small neighborhood of the point $v=1$. If $0 < \delta < 1$ then $\int_0^\delta e^{n[v+\log v]}\,dv$ is exponentially smal when compared to $\int_\delta^1 e^{n[v+\log v]}\,dv$. Oct 4 '15 at 13:48