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Set $R_n(x)$ the quantity in the title, for $n\in\mathbb{N}$ and $x > 0$.

I'm trying to prove that if $x < 1$, then $R_n(x)\underset{n\rightarrow +\infty}{\longrightarrow} 0$, if $x > 1$, then $R_n(x)\underset{n\rightarrow +\infty}{\longrightarrow} 1$ and if $x = 1$, then $R_n(x)\underset{n\rightarrow +\infty}{\longrightarrow} \frac{1}{2}$. (I got these limits with Python.)

My approach is to make a substitution in the integral, $u = nt$. I get then $R_n(x) = \displaystyle\frac{1}{n!}\int_0^{nx}e^{-u}u^n\,\mathrm du$. I recognize the (incomplete) Gamma function, and know that $\Gamma(n+1) = n!$ but don't know how to proceed (I tried many other substitutions, without success).

Could I get some help on the matter? Thanks.

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2 Answers 2

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Here's a non-probabilistic casting of @user10354138's logic.

You want $$\lim_{n\to\infty}\frac{n^{n+1}}{n!}\int_0^x e^{-nt}t^ndt=\lim_{n\to\infty}\frac{1}{n!}\int_0^{nx}e^{-y}y^ndy.$$A bit of calculus gvies the quadratic approximation $n\ln y-y\approx n\ln n-n-\frac{(y-n)^2}{2n}$, so$$\int_0^{nx}e^{-y}y^ndy\approx n^ne^{-n}\int_0^{nx}e^{-(y-n)^2/(2n)}dy\approx\frac{n!}{\sqrt{2\pi n}}\int_{-n}^{n(x-1)}e^{-z^2/(2n)}dz,$$where the second $\approx$ uses the Stirling approximation. So we want$$\lim_{n\to\infty}\frac{1}{\sqrt{2\pi n}}\int_{-n}^{n(x-1)}e^{-z^2/(2n)}dz.$$If $x>1$, the integral $\approx\int_{-\infty}^\infty e^{-z^2/(2n)}dz=\sqrt{2\pi n}$, so the limit is $1$. If $x=1$, we lose the $z>0$ half of the integral, halving the result; if $x<1$, the upper limit $\to-\infty$, making the limit $0$.

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A probabilistic proof: Apply the central limit theorem to independent exponentially distributed random variables $X_1,X_2,\dots\sim\operatorname{Exp}(\lambda=1)$. Note that $X_1+\dots+X_n\sim \operatorname{Gamma}(\alpha=n, \beta=1)$ has mean $n$ and variance $n$. So $$ R_n(x)=\mathbb{P}(X_1+\dots+X_n\leq nx)\simeq\Phi((x-1)\sqrt{n})\to\begin{cases} 0 & x<1\\ 1 & x>1\\ \frac12 & x=1 \end{cases}. $$


If you don't like probability, you can directly appeal to asymptotic expansions such as $$ \frac{\gamma(n+1,a+y(2n)^{1/2})}{\Gamma(n+1)} =\frac12\operatorname{erfc}(-y)-\frac13\left(\frac2{n\pi}\right)^{1/2}(1+y^2)\exp(-y^2)+O(n^{-1})\quad (y\in\mathbb{R}, n\to\infty) $$ found by Tricomi in 1950.

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  • $\begingroup$ Thanks, but I am not that well versed in probability :/ Do you have a more analytic proof? $\endgroup$
    – Nathaniel
    Jun 2, 2019 at 15:36

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