Using one of the Lebesgue convergence theorems, find $$\lim_{n \to \infty}\int_{[0,n]}x\left(1-\frac{x}{n}\right)^{n} dm(x)$$ I'm not sure how to show that the sequence of functions is increasing, or how to find the limit of them, in order to be able to apply the extended monotone convergence theorem.
I'd appreciate a point in the right direction.
I had thought about using the dominated convergence theorem, where
$\left|x\left(1-\frac{x}{n}\right)^{n}\right| = \left|x\right|\left|\left(1-\frac{x}{n}\right)^{n}\right|\leq \left|x\right|$ as $x \in [0,n]$ so $\left(1-\frac{x}{n}\right)\leq1$
But as the integral is over $[0,n]$, im not sure if the function $x$ works with the dominated convergence theorem, as over $[0,\infty]$ the integral of $x$ is clearly infinite.
It is easy to show $\int_{[0,n]}x\left(1-\frac{x}{n}\right)^{n}dm(x)$ converges. Define $$ f_n(x)=x\left(1-\frac{x}{n}\right)^{n}\chi_n(x) $$ and hence $$\int_{[0,n]}x\left(1-\frac{x}{n}\right)^{n} dm(x)=\int_{[0,\infty)} f_n(x)dm(x). $$ Let $g_n(x)=x^3\left(1-\frac{x}{n}\right)^{n}$ for $x\in[1,n]$. It is easy to see $g_n(x)$ attains the max $\frac{37n^3}{(n+3)^3}(\frac n{n+3})^n$ at $x=x_n:=\frac{3n}{n+3}$. Since $g_n(x_n)\to \frac{37}{e^3}$ as $n\to\infty$, one has $$ 0\le f_n(x)=\frac{1}{x^2}g_n(x)\chi_n(x)\le \frac{C}{x^2}, x\ge 1$$ for some constant $C>0$. Since $\int_1^\infty\frac{C}{x^2}dx $ converges, by the DCT, $$ \lim_{n\to\infty}\int_{[1,\infty)} f_n(x)dm(x)=\int_{[0,\infty)} xe^{-x}dm(x). $$