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$\qquad\qquad\qquad$ 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 ?$)$

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Lucian
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  • 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.. – tired Oct 04 '15 at 12:19

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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

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

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Antonio Vargas
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  • nice (+1), i was working on the same answer but you beat me for approximatly five minutes! Gratulations! :) – tired Oct 04 '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? – Calvin Khor Oct 04 '15 at 11:45
  • @tired http://i.minus.com/iiD0TlBgErfpi.gif – Antonio Vargas Oct 04 '15 at 11:49
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    @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. – Antonio Vargas Oct 04 '15 at 11:50
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    @AntonioVargas http://theartmad.com/wp-content/uploads/2015/08/Lucky-Luke-Shadow-2.jpg ;) – tired Oct 04 '15 at 11:50
  • @AntonioVargas Thanks :) – Calvin Khor Oct 04 '15 at 11:55
  • I believe that the Laplace method argument could be made more rigorous by squeezing, since $2x-1$ isn't a very good approximation of $x+\ln x$ near $0$, but, either way, $(+1)$. – Lucian Oct 04 '15 at 12:45
  • @Lucian, one proof of this version of the Laplace method (de Bruijn, p. 65, which outlines an argument to follow the framework of an earlier section) definitely does involve squeezing as you suggest. – Antonio Vargas Oct 04 '15 at 12:48
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    @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 $\int_\delta^1 e^{n[v+\log v]},dv$. – Antonio Vargas Oct 04 '15 at 13:48