I know the Law of Iterated Logarithms states the following almost surely:
$$\limsup_{t\to\infty} \frac{B(t)}{\sqrt{2t\log\log t}} = 1 $$
I was wondering if there are similar ones. For example, if I was interested in:
$$\limsup_{t\to\infty} \frac{B(t)}{\log(t)}$$
Is there a way to tell if the above expression converges to a constant $k$ or if it grow to $\infty$ a.s.? Or for example if I had an arbitrary continuous function $\phi(t)$ instead of $\log(t)$, can one determine if the $\limsup$ is finite or infinite?
Assume that $g\colon \mathbf R_+\to\mathbf R_+^*$ is a function such that $$\lim_{t\to\infty}\frac{g(t)}{\sqrt{t\log\log t} } =0.$$ Then we have $$\limsup_{t\to +\infty }\frac{B(t) }{g(t)}=+\infty.$$ Indeed, since $\limsup_{t\to\infty} \frac{B(t)}{\sqrt{2t\log\log t}} = 1$, we have $$\lim_{R\to \infty}\sup_{t\geqslant R}\frac{B(t)}{\sqrt{2t\log\log t}}=1,$$ hence, for almost every $\omega$, one can find $R(\omega)$ such that if $t\geqslant R(\omega)$, then $$\frac 12\leqslant \frac{B(t)}{ \sqrt{2t\log\log t} }\leqslant \frac 32,$$
from which it follows that for $t\geqslant R(\omega)$, $$\frac 12 \frac{\sqrt{2t\log\log t}}{g(t)} \leqslant \frac{B(t)}{g(t)}.$$