Let $X_1, X_2, ...$ be a sequence of random variables (finite almost surely), and let $0 < b_n \uparrow \infty$. As part of solving a problem I seem to have shown that $\frac{X_n}{b_n} \to 0$ almost surely implies $\frac{\max_{1 \le j \le n} |X_n|}{b_n} \to 0$ almost surely. I want to confirm that this is true since, as it pertains to the full problem I am solving, it doesn't use all of the hypotheses I am given. It seems simple enough, but I'm wondering if I am making a mistake somewhere.
Fix $n > k \ge 2$. Then $\frac{\max_{1 \le j \le n} |X_j|}{b_n} \le \frac{\max_{1 \le j \le k} |X_j|}{b_n} + \frac{\max_{k \le j \le n} |X_j|}{b_n} \le \frac{\max_{1 \le j \le k} |X_j|}{b_n} + \max_{k \le j \le n} \frac{ |X_j|}{b_j}$
the last inequlaity because $b_n \uparrow$. Letting $n \to \infty$ gives $\limsup_n\frac{\max_{1 \le j \le n} |X_j|}{b_n} \le \sup_{j \ge k} \frac{|X_j|}{b_j}$ almost surely and letting $k \to \infty$ gives $\limsup_n\frac{\max_{1 \le j \le n} |X_j|}{b_n} \le \limsup_n \frac{|X_n|}{b_n}$
and by hypothesis the term on the RHS is equal to $0$ almost surely.
Is this okay or are there holes?