Suppose $B_t$ is a standard Brownian motion. Does there exist $c < \infty$ such that with probability one$$\limsup_{t \to \infty} {{B_t}\over{\sqrt{t \log t}}} \le c?$$I need to know whether or not this is true for something I am doing, but I do not do probability by trade...

4$\begingroup$ google "law of iterated logarithm" to see that the answer is positive with c=0 $\endgroup$– ofer zeitouniAug 5 '15 at 1:27
As already stated in the comments it follows from the law of iterated logarithm for Brownian motion.
Write
$\displaystyle{\limsup_{t\rightarrow \infty} \frac{B_t}{\sqrt{t \log(t)}} = \limsup_{t\rightarrow \infty} \frac{B_t}{\sqrt{2t\log(\log(t))}} \frac{\sqrt{2\log(\log(t))}}{\sqrt{\log(t)}}} = 0$
as $\displaystyle{\limsup_{t\rightarrow \infty} \frac{B_t}{\sqrt{2t\log(\log(t))}} = 1}$ a.s. by the law of iterated logarithm and $\displaystyle{\lim_{t\rightarrow \infty}\frac{\sqrt{2\log(\log(t))}}{\sqrt{\log(t)}} = 0}$, which can be easily shown by L'Hopital's rule.
Therefore, the optimal constant is $C=0$.