Limit of Riemannian integrals with Lebesgue

Question

Show: $\lim\limits_{n\rightarrow\infty}\int_0^nx^i(1-\frac{x}{n})^ndx=i!$ for $i\geq 0$.

Hello there. Question is how do this? Which relation between Lebesgue and Riemann integrals must be used? Appreciate any hints. Thank you.

Answer 1

It should be $$ 0\lt\int_0^n x^i\left(1-\frac{x}{n}\right)^ndx\le \int_0^n x^ie^{-x}dx\le\int_0^\infty x^ie^{-x}dx=\Gamma(i+1)=i!. $$

Answer 2

Let $$F(i,k)=\int_{0}^{n}x^i\Big(1-\frac{x}{n}\Big)^k\ dx,$$ then integration by parts gives $$F(i,k)=\frac{in}{k+1}F(i-1,k+1)\text{ and }F(0,k)=\frac{n}{k+1}\ .$$

Thus, $$ \begin{array}{rcl} F(i,n)&=&\displaystyle{\frac{ni}{n+1}F(i-1,n+1)}\\ &=&\displaystyle{\frac{i(i-1)n^2}{(n+1)(n+2)}F(i-2,n+2)}\\ &\vdots&\\ &=&\displaystyle{\frac{i!n^i}{(n+1)(n+2)\cdots(n+i)}F(0,n+i)}\\ &=&\displaystyle{(i!)\times\underbrace{\frac{n^{i+1}}{(n+1)(n+2)\cdots(n+i+1)}}_{\to 1\text{ as }n\to\infty}} \end{array}$$

Thus $F(i,n)\to i!$ as $n\to\infty$.

Now, if you're worried about the integrals converging, you can use the dominated convergence theorem.