Let $(B_t)_{t\geq 0}$ be brownian motion, let $p>1/2$. I want to show that $$\lim_{t\to\infty} \frac{B_t}{t^p} \to 0 \quad a.s.$$
Atm I'm trying to show that $$\limsup_{t\to \infty} \frac{B_t}{t^p} \to 0 \quad a.s.$$
by applying Borel-Cantelli on the event $A = \limsup_{n \to \infty}A_n, A_n = \{\sup_{n-1\leq t \leq n} B_t > C(n-1)^p \}, C > 0.$ This means I want to show that $\sum_{n=1}^\infty P(A_n) < \infty$, which implies $P(A) = 0 \quad a.s. \Rightarrow \limsup_{t\to \infty} \frac{B_t}{t^p} \to 0 \quad a.s.$. By the reflection principle $$P(\sup_{n-1\leq t \leq n} B_t > C(n-1)^p) \leq P(\sup_{0\leq t \leq n} B_t > C(n-1)^p)$$ $$=2P(B_n > C(n-1)^p) = 2P(B_1 > C\frac{(n-1)^p}{\sqrt{n}}).$$ So my question is: How can I show that $$\sum_{n=1}^\infty P(B_1 > C\frac{(n-1)^p}{\sqrt{n}}) < \infty$$ for $C>0,p>1/2, B_1 \sim N(0,1)$?
Btw: i'm not even sure whether the sum converges, maybe i have to pick other intervals and not $[n-1,n]$.
Since $p>\frac{1}{2}$, we can write $p= \frac{1}{2}+q$ for some $q>0$. Now
$$\frac{(n-1)^p}{\sqrt{n}} = \sqrt{\frac{n-1}{n}} (n-1)^q \geq \frac{1}{2} (n-1)^q, \qquad n \in \mathbb{N},$$
implies
$$\mathbb{P} \left(B_1 > C \frac{(n-1)^p}{\sqrt{n}} \right) \leq \mathbb{P} \left(B_1 > \frac{C}{2} (n-1)^q\right).$$
Choose $k \in \mathbb{N}$ sufficiently large such that $qk>1$. By the Markov inequality,
$$\mathbb{P}\left(B_1 > \frac{C}{2} (n-1)^q\right) \leq C' \frac{1}{(n-1)^{qk}} \mathbb{E}(|B_1|^k)$$
for $C' := \left( \frac{2}{C} \right)^k$. (Note that we used here that a Gaussian random variable has moments of arbitrary order.) Combining both estimates yields
$$\sum_{n \geq 2} \mathbb{P} \left(B_1 > C \frac{(n-1)^p}{\sqrt{n}} \right) \leq C' \mathbb{E}(|B_1|^k) \sum_{n \geq 2}\frac{1}{(n-1)^{qk}}< \infty.$$