$\limsup \frac{X_n}{a_n}=1 \Leftrightarrow \limsup \frac{\max\{X_1,...,X_n\}}{a_n}=1$ $\{X_n\}_{n=1}^\infty$ is non-negative random variables, $0\leq a_n \uparrow\infty$. Then:
\begin{equation}
\limsup \frac{X_n}{a_n}=1 \quad a.s.\Leftrightarrow \limsup \frac{\max\{X_1,...,X_n\}}{a_n}=1 \quad a.s.
\end{equation}
My ideas so far:
Since $X_n$ and $a_n$ are non-negative, we have:
\begin{equation}
0\leq \frac{X_n}{a_n}\leq \frac{\max\{X_1,...,X_n\}}{a_n}
\end{equation}
So, if $\limsup \frac{\max\{X_1,...,X_n\}}{a_n}=1$, then $\limsup\frac{X_n}{a_n}\leq 1.$
But I cannot going on to prove $\limsup\frac{X_n}{a_n}\geq 1.$
Thanks in advance for any help!
 A: You have in fact shown that for any deterministic sequence $(X_n)$ we have
$$\limsup \frac{X_n}{a_n} \leq \limsup \frac{\max (X_1,\ldots,X_n)}{a_n}.$$
Now if we show that (for $(X_n)$ deterministic)
\begin{equation}\limsup \frac{\max (X_1,\ldots,X_n)}{a_n} \leq \limsup \frac{X_n}{a_n},\end{equation}
then it means that the limits are equal. Therefore your statement follows because the limits are equal for every fixed $\omega$ in the probability space.
Let $\epsilon >0$ and assume that $\limsup \frac{X_n}{a_n}$ is finite, say equal to $1$ (if it's infinite, then both limits are infinite by the inequality that you have shown). Then there exists $N$ such that for all $n\geq N$ we have $$\frac{X_n}{a_n} < 1+\epsilon.$$ We can also assume that $a_{N}\geq 1$. Since $a_n \uparrow \infty$, there exists $M$ such that $$a_{N+M}\geq \max (X_1,\ldots,X_{N-1})a_N.$$  Therefore, if $k\geq N+M$, the monotonicity of $(a_n)$ gives
\begin{align*}
\frac{\max(X_1,\ldots,X_k)}{a_k}&\leq \max\bigg(\frac{\max(X_1,\ldots,X_{N-1})}{a_k}, \frac{\max(X_N,\ldots,X_{k})}{a_k}\bigg)\\
&\leq \max\bigg(\frac{\max(X_1,\ldots,X_{N-1})}{\max (X_1,\ldots,X_{N-1})a_N}, \max\big(\frac{X_N}{a_N},\frac{X_{N+1}}{a_{N+1}},\ldots,\frac{X_{k}}{a_k}\big)\bigg)\\
&\leq \max\big(\frac 1{a_N}, 1+ \epsilon\big) = 1+\epsilon.
\end{align*}
This proves our inequality.
