For every $\epsilon>0$, the probability of $W_t>(1+\epsilon)\sqrt{t\log(t)}$ tends to $0$ as $t\to\infty$ Can anybody give a hint to show for all $\epsilon>0$
$$\lim_{t \to \infty} P \left( \frac{W_t}{\sqrt{t\log(t)}}>1+\epsilon \right) = 0$$
with $W_t$ Brownian Motion? (Or W(t), a Brownian motion on time t that behaves as a random walk)
I tried to use that $\frac{e^{W_t^2/(1+2t)}}{\sqrt{1+2t}}$ is a Martingale. I think that is useful, but I get stuck, that's why I come here for help. Any useful theorems or methods?
 A: With $W_t$ Brownian motion, we have (for each $t$):
$$
\begin{eqnarray*}
\mathbb{E}\left[\frac{W_t}{\sqrt{t\log(t)}}\right]&{}={}&0\,,\newline
\mathbb{V}ar\left[\frac{W_t}{\sqrt{t\log(t)}}\right]&{}={}&\frac{1}{\log\left(t\right)}\,.
\end{eqnarray*}
$$
Consequently, using Chebyshev's Inequality, we have (for each $t$):
$$
P \left( \frac{W_t}{\sqrt{t\log(t)}}>1+\epsilon \right){}\leq{} P \left( \left|\frac{W_t}{\sqrt{t\log(t)}}\right|>1+\epsilon \right){}\leq{}\frac{\mathbb{V}ar\left[\frac{W_t}{\sqrt{t\log(t)}}\right]}{\left(1+\epsilon\right)^2}{}={}\frac{1}{\log(t)\left(1+\epsilon\right)^2}\,.
$$
Therefore,
$$
0{}\leq{}\lim_{t \to \infty}\,P \left( \frac{W_t}{\sqrt{t\log(t)}}>1+\epsilon \right){}\leq{}\lim_{t \to \infty}\,\frac{1}{\log(t)\left(1+\epsilon\right)^2}{}={}0\,.
$$
A: If $W_t$ is a standard Brownian motion, then for a fixed $t$, $W_t$ is normally distributed with variance $t$ and mean zero. We thus have to show that  $$p(t):=\mathbb P\left(W_1\gt (1+\varepsilon)\sqrt t\right)$$
converges to $0$ as $t$ goes to infinity. This is due to the fact that if $X$ is a random variable and $h\colon\mathbb R_+\to\mathbb R_+$ is a function such that $\lim\limits_{t\to +\infty}h(t)=+\infty$, then $\lim\limits_{t\to +\infty}\mathbb P\left(X\gt h(t)\right)=0$. To show that, use the fact that the sequence of sets $\left(\{X\gt n\} \right)_{n\geqslant 1}$ is non-increasing. 
