Computing the limit $\lim_{k \to \infty} \int_0^k x^n \left(1 - \frac{x}{k} \right)^k \mathrm{d} x$ for fixed $n \in \mathbb{N}$

Question

I'm working on a problem that asks to compute $$\lim_{k \to \infty} \int_0^k x^n \left(1 - \frac{x}{k} \right)^k \mathrm{d} x$$ for fixed $n \in \mathbb{N}$. What I've tried so far is to do a $u$-substitution for $u = \frac{x}{k}$, so I have $$\int_0^k x^n \left(1 - \frac{x}{k} \right)^k \mathrm{d} x = k^{n + 1} \int_0^1 u^n (1 - u)^k \mathrm{d} u .$$ Using the Binomial Theorem to break up the $(1 - u)^k$ term, I get $$k^{n + 1} \int_0^1 u^n (1 - u)^k \mathrm{d} u = k^{n + 1} \int_0^1 \sum_{j = 0}^k \binom{k}{j} \frac{(-1)^j}{n + j + 1} .$$ However, I don't know how to compute the limit of this expression as $k \to \infty$. I assume that I should recognize it as some kind of Taylor series that's somehow $O \left( k^{-(n + 1)} \right)$, but I'm not seeing it.

Note: When looking at other posts, I found an integral that looked similar to this one, and the only answer on that post involved something called a beta function. I have never heard of a beta function, and would like to find a solution here that doesn't rely on whatever a beta function is.

Another idea I considered was to use the Dominated Convergence Theorem, since $e^{-x} = \lim_{k \to \infty} \left( 1 - \frac{x}{k} \right)^k$, so I figured I could use DCT to say that \begin{align*} \lim_{k \to \infty} \int_0^k x^n \left( 1 - \frac{x}{k} \right)^k \mathrm{d} x & = \lim_{k \to \infty} \int_0^\infty \chi_{[0, k]}(x) x^n \left( 1 - \frac{x}{k} \right)^k \mathrm{d} x \\ & = \int_0^\infty x^n e^{-x} \mathrm{d} x & (\textrm{DCT used here})\\ & = n ! , \end{align*} assuming I didn't mess up any of my integration by parts. However, I couldn't find a choice of dominator that would work on all of $[0, \infty)$, so I'd also be interested in a solution that uses DCT as well. Perhaps I'm being naive, but the pointwise limit is just so convenient that I have to imagine that DCT can be used here.

EDIT: Thanks to some inspiration from a comment by user Mars Plastic, I thought to consider some other integral convergence theorems. I came up with this, which I think works. I'm still interested to see if the Dominated Convergence Theorem argument can be made to work, perhaps a bit more smoothly than this.

Let $f_k(x) = \chi_{[0, k]}(x) x^n \left( 1 - \frac{x}{k} \right)^k$. I claim that the sequence is monotone increasing on $[0, \infty)$. Fix $x \in (0, \infty)$, and let $K = \lfloor k \rfloor + 1$, so that $K = \min \{ k \in \mathbb{N} : f_k(x) \neq 0 \}$. Obviously if $k < K$, then $f_k(x) = 0 \leq f_{k + 1}(x)$. So consider the case where $k \geq K$. Then $f_k(x) = x^n \left(1 - \frac{x}{k} \right)^k$, and based on answers to this question, it seems this sequence would be monotone increasing. Therefore, I can apply the Monotone Convergence Theorem to say that $f_k(x) \nearrow x^n e^{-x}$, so $$\int_0^k x^n \left( 1 - \frac{x}{k} \right)^k \mathrm{d} x = \int_0^\infty f_k(x) \mathrm{d} x = \int_0^\infty x^n e^{-x} = n! .$$

Answer 1

To apply Lebesgue's dominated convergence theorem, it suffices to show that $\left( 1 - \frac{x}{k} \right)^k \leq e^{-x}$ on $(0,k)$, because, if so, we have the uniform bound $$x^n \left( 1 - \frac{x}{k} \right)^k \chi_{(0,k)} \leq x^ne^{-x} $$ on $(0, \infty)$, for all $n$. Since $x^ne^{-x}$ is integrable, the conditions of LDCT would be met.

Now, observe our claim is equivalent to showing that $f(x) = k \log \left( 1 - \frac{x}{k} \right) + x \leq 0$ for $ x\in (0,k)$. Note that the inequality holds for $f(0) = 0$. Also, $$f^\prime (x) = 1-\frac{1}{1 - \frac{x}{k}} = -\frac{x/k}{1-x/k} < 0 \qquad x \in (0,k)$$ Thus, $f$ is decreasing on $(0,k)$, and $f(x) \leq 0$ on $x \in (0,k)$, as desired.

Answer 2

Here is a quick rundown on a Beta function, as it is useful in this case but also good to know in general: $$\begin{align} \Gamma(x)\Gamma(y)=&\int_0^\infty u^{x-1}e^{-u}\,du\int_0^\infty v^{y-1}e^{-v}\,dv\\ =&\int_0^\infty\int_0^\infty u^{x-1}v^{y-1}e^{-(u+v)}\,du\,dv \end{align}$$ then using $u=zt,v=z(1-t)$ gives: $$\begin{align} \phantom{a}&\int_0^\infty\int_0^1e^{-z}(zt)^{x-1}(z(1-t))^{y-1}z\,dt\,dz\\ \phantom{a}&\underbrace{\int_0^\infty e^{-z}z^{(x+y)-1}\,dz}_{\Gamma(x+y)}\cdot\underbrace{\int_0^1t^{x-1}(1-t)^{y-1}\,dt}_{B(x,y)} \end{align}$$ and so it follows that: $$B(x,y)=\int_0^1t^{x-1}(1-t)^{y-1}\,dt=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$ where $\Gamma(n)=(n-1)!$

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In your case: $$k^{n+1}\int_0^1u^n(1-u)^k\,du=k^{n+1}B(n+1,k+1)=k^{n+1}\frac{n!k!}{(n+k+1)!}$$

Answer 3

Let $\displaystyle I(n, k) = \int_0^1 u^n (1 - u)^k \,\mathrm{d} u$. Then integrating by parts:

$\begin{aligned} I(n, k) & = \frac{1}{n+1}\int_0^1 (u^{n+1})' (1 - u)^k \,\mathrm{d} u \\& = \frac{1}{n+1}\cdot u^{n+1}(1-u)^{k}\bigg|_0^1-\frac{1}{n+1}\int_0^1 u^{n+1}[(1-u)^{k}]'\, \mathrm du \\& = \frac{k}{n+1}\int_0^1 u^{n+1}(1-u)^{k-1}\, \mathrm du \\& = \frac{k}{n+1} I(n+1, k-1). \end{aligned}$

Carrying out the recurrences:

$\begin{aligned} I(n, k) & = \frac{k}{n+1} I(n+1, k-1) \\& = \frac{k}{n+1}\cdot \frac{k-1}{n+2} \cdots \frac{1}{n+k}I(n+k, 0) \\& = \frac{k}{n+1}\cdot \frac{k-1}{n+2} \cdots \frac{1}{n+k}\int_0^{1} u^{n+k} \, \mathrm du \\& = \frac{k}{n+1}\cdot \frac{k-1}{n+2} \cdots \frac{1}{n+k} \cdot \frac{1}{n+k+1} \\& = \frac{k!n!}{(k+n+1)!}. \end{aligned} $

Therefore

$$I = \lim_{k \to \infty} \frac{k!n! k^{n+1}}{(k+n+1)!} = n! \cdot \lim_{k \to \infty} \frac{k! k^{n+1}}{(k+n+1)!} = n!$$