I have tried to solve the following exercise: let $B=(\Omega,\mathcal{F},\{\mathcal{F}_t\}_{t\in T},\{X_t\}_{t\in T},P)$ and $W=(\Omega,\mathcal{F},\{\mathcal{F}_t\}_{t\in T},\{W_t\}_{t\in T},P)$ two independent browinan motion. Compute \begin{equation} \limsup_{t\to\infty}|B_t|e^{|W_t|} \end{equation}
MY ATTEMPT: My idea is to show that $\limsup_{t\to\infty}|B_t|=\infty$ and this is sufficient to conclude that also the above $\limsup$ is $\infty$ since $e^{|W_t|}\geq 1$ for all $t\in T$. I now consider $\{t_n\}_n$ an increasing sequence to $\infty$, $\delta>0$ a parameter and I let the event: \begin{equation} A_n=\{\omega\in\Omega:|B_{t_n}|<2^n(1+\delta)\} \end{equation} Fix now $t_n:=2^{4n}$, then we want to compute the following: \begin{equation} P(A_n)=P(|B_{2^{4n}}|<2^n(1+\delta))=P\bigg(\bigg|\frac{B_{2^{4n}}}{2^n}\bigg|<1+\delta\bigg)=\int_{-(1+\delta)}^{1+\delta}\frac{1}{2^n\sqrt{2\pi}}\exp(-x^2/2^{2n+1})dx\leq \frac{C}{2^n} \end{equation} Where we used that $\frac{B_{2^{4n}}}{2^n}\sim N(0,2^{2n})$. Now this last is the general term of a convergent series and allow us to prove that $\sum_n P(A_n)<\infty$ which implies that $P(\lim_n\sup A_n)=0\Rightarrow P(\lim_n\inf A_n^c)=1$ which implies that almost surely for all $\delta >0$: \begin{equation} \limsup_{n\to\infty}\bigg|\frac{B_{t_n}}{2^n}\bigg|\geq (1+\delta) \ \ \ \text{a.s.} \end{equation}
I'm not completely sure on the proof and especially on the last computations, can someone help me to complete correctly this proof?