Limit of an integral related to Gamma function: $ \lim_{k \to \infty}\int_{0}^{k} x^{n}\left(1 - {x \over k}\right)^{k}\,\mathrm{d}x = n! $

Question

I have the following equality: $$ \lim_{k \to \infty}\int_{0}^{k} x^{n}\left(1 - {x \over k}\right)^{k}\,\mathrm{d}x = n! $$

What I think is that after taking the limit inside the integral ( maybe with the help of Fatou's Lemma, I don't know how should I do that yet ), then get

$$ \int_{0}^{k}\lim_{k \to \infty}\left[\,% x^{n}\left(1 - {x \over k}\right)^{k}\,\right]\mathrm{d}x = \int_{0}^{\infty}x^{n}\,\mathrm{e}^{-x}\,\mathrm{d}x = \Gamma\left(n + 1\right) = n! $$

How can I give a clear proof ?.

Answer 1

By writing $$ \int_{0}^{k} x^{n}\Big(1-\cfrac{x}{k}\Big)^{k}dx=\int_{0}^{\infty}\mathbf{1}_{[0,k]}(x)\: x^{n}\Big(1-\cfrac{x}{k}\Big)^{k}dx, $$ observing that $$ 0\le \mathbf{1}_{[0,k]}(x)\: x^{n}(1-\cfrac{x}{k})^{k}\le x^ne^{-x}, \quad x\ge0, $$ and, for $x\ge 0$, $$ \lim_{k \rightarrow \infty}\Big(1-\cfrac{x}{k}\Big)^{k}=e^{-x} $$ one may use the Dominated Convergence Theorem.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Besides the short $\texttt{@Olivier Oloa}$ answer, another straightforward alternative is based on the Beta Function as follows:

\begin{align} \lim_{k \to \infty}\int_{0}^{k} x^{n}\pars{1 - {x \over k}}^{k}\,\dd x & \,\,\,\stackrel{x/k\ \mapsto\ x}{=}\,\,\, \lim_{k \to \infty}\bracks{k^{n + 1}\int_{0}^{1}x^{n}\pars{1 - x}^{k}\,\dd x} \\[5mm] & = \lim_{k \to \infty}\bracks{k^{n + 1}\,\mrm{B}\pars{n + 1,k + 1}}\qquad \pars{~\mrm{B}:\ Beta\ Function~} \\[5mm] & = \lim_{k \to \infty}\bracks{k^{n + 1}\, {\Gamma\pars{n + 1}\Gamma\pars{k + 1} \over \Gamma\pars{k + n + 2}}}\qquad \pars{~\Gamma:\ Gamma\ Function~} \\[5mm] & = n!\,\lim_{k \to \infty}\bracks{k^{n + 1}\, {k! \over \pars{k + n + 1}!}} \\[5mm] & = n!\,\lim_{k \to \infty} {1 \over \bracks{1 + \pars{n + 1}/k}\bracks{1 + n/k}\ldots\pars{1 + 1/k}} = \bbx{\ds{n!}} \end{align}