Let $X_1,X_2,...$ be positive integer-valued random variables with the same distribution but they can be dependent on each other. We want to prove that $\lim_{n\rightarrow \infty} E[\max_{i<n} X_i/n] = 0$.
I keep running into issues with the lack of independence. I want to do the classic "if max is less than $\epsilon$ then they all have to be so raise the probability of one to the $n$" type of trick, but it obviously fails if they are dependent. Maybe the idea can be used, but not sure how.
Clearly $$E\left[ \max_{i<n} \frac{X_i}{n} \right] = \sum_{m=0}^n P\left(\max_{i<n} \frac{X_i}{n}>m \right)$$ which seems like it should be the key, but I keep hitting a wall on bounding this above in a way that allows the limit to work.
This is from Ross and Pekoz "A Second Course in Probability" Chapter 1, Problem 12. Just working on it as a summer project for myself.
As pointed out by LostStatistician18, you have to assume that $\mathbb E[X_1]$ is finite, otherwise, the expectation into question is infinite for each $n$.
Start from $$ \frac 1n\max_{1\leqslant i\leqslant n}X_i\leqslant \frac 1n\max_{1\leqslant i\leqslant n}X_i\mathbf{1}_{\{X_i\leqslant R\}}+\frac 1n\max_{1\leqslant i\leqslant n}X_i\mathbf{1}_{\{X_i> R\}}\leqslant \frac Rn+\frac 1n\sum_{i=1}^n X_i\mathbf{1}_{\{X_i> R\}},$$ valid for each positive $R$.
Take the expectation and use the fact that the random variable $X_i\mathbf{1}_{\{X_i> R\}}$ have the same expectation to get $$ \mathbb E\left[\frac 1n\max_{1\leqslant i\leqslant n}X_i\right]\leqslant \frac Rn+\mathbb E\left[X_1\mathbf{1}_{\{X_1> R\}}\right]. $$