Let $(B_t)_{t\geq 0}$ be a brownian motion, i want to show that $$\frac{B_t}{t^p} \xrightarrow[t\to\infty]{a.s.} 0, $$ for all $p>\frac{1}{2}$.
I was told to use that
$$X_t = \frac{B_t^2 - t}{(t+1)^{2p}} \xrightarrow[t\to\infty]{a.s.} 0,$$
which i have proven already, but i can't see how $X_t \to 0 \, a.s.$ implies the first claim. Can someone give me a hint on how those two are related?
If $p>1/2$ then $\frac{t}{(t+1)^{2p}} \to 0$. Combined with $X_t \to 0$, this implies $X_t + \frac{t}{(t+1)^{2p}} \to 0$. But $$X_t + \frac{t}{(t+1)^{2p}} = \frac{B_t^2}{(t+1)^{2p}} = \left(\frac{B_t}{(t+1)^p}\right)^2 \to 0.$$ It follows that $$ \frac{B_t}{(t+1)^p} \to 0 $$ and thus $$ \frac{B_t}{t^p} = \frac{B_t}{(t+1)^p} \frac{(t+1)^p}{t^p} \to 0 \cdot 1 = 0 $$