If $X_1,X_2,...$ are non-negative random variables with the same distribution (but the variables are not necessarily independent) and $E[X_1]< \infty $, prove that $$\lim_{n \to \infty}E\left[\max_{ i< n}\frac{X_i}{n}\right]=0.$$
This is what I have so far
Xi = √n+(Xi −√n)
≤√n+(Xi −√n)+(=max)
maxi< n Xi ≤ √n +(=max)
maxi< n(Xi − √n)+(=max)
≤√n + sigma from i<n of (Xi −√n)+(=max)
how do I finish this to establish the desired result
Let $M_n=\max\{X_i\mid i\leqslant n\}$ and $u_n(x)=\frac1nP(M_n\geqslant x)$ then $$\frac{E[M_n]}n=\int_0^\infty u_n(x)\mathrm dx.$$ Note that $u_n\to0$ pointwise since $u_n\leqslant\frac1n$ uniformly and that, for every $x$, $[M_n\geqslant x]\subseteq\bigcup\limits_{i\leqslant n}[X_i\geqslant x]$ hence $P(M_n\geqslant x)\leqslant nP(X_1\geqslant x)$, that is, $u_n\leqslant u_1$.
Since $u_1$ is integrable, Lebesgue dominated convergence theorem then shows that $$\int_0^\infty u_n(x)\mathrm dx\to0.$$