Show $\limsup _{t \to \infty} \frac{B_t}{\sqrt{t \ln t}} \leq 1 $ using the fact that $\frac{e^{B_t ^2 / (1+2t)}}{\sqrt{1+2t}}$ is a martingale. For a standard Brownian motion $(B_t)_{t \geq 0}$, I want to show that $$\limsup _{t \to \infty} \frac{B_t}{\sqrt{t \ln t}} \leq 1 \ \ \text{a.s.}$$ using the fact that $\frac{\exp(B_t ^2 / (1+2t))} { \sqrt{1+2t}}$ is a martingale.
It suffices to show $\limsup _{t \to \infty} \frac{B_t ^2}{t \ln t} \leq 1 \ \ \text{a.s.}$ and I know that $\frac{B_t ^2}{t \ln t} \leq \frac{B_t ^2} {1+2t}$ for large $t$, but I don't see any further.
Any help is appreciated.
 A: Thanks to @John Dawkins, I have come up with an answer.
By the martingale convergence theorem, our martingle converges pointwise to a non-negative real valued random variable $Z$ a.s. Note that
$$\frac {e^{B_t^2/(1+2t)}}{\sqrt{1+2t}} = \exp \bigg({\frac{t \ln t}{1+2t} \left( \frac{B_t^2}{t \ln t}- \frac{(1+2t)\ln(1+2t)}{2t \ln t}\right) \bigg)}$$
On the event $Z > 0$ (and where the martingale converges to $Z$, of course)
$$\lim _{t \to \infty}\frac{t \ln t}{1+2t} \left( \frac{B_t^2}{t \ln t}- \frac{(1+2t)\ln(1+2t)}{2t \ln t}\right) \bigg) \in \mathbb R$$ 
which implies $\lim_{t \to \infty} \frac{B_t ^2}{t \ln t} = 1$. On the event $Z=0$
$$\lim _{t \to \infty}\frac{t \ln t}{1+2t} \Big( \frac{B_t^2}{t \ln t}- \frac{(1+2t)\ln(1+2t)}{2t \ln t}\Big) \bigg) = -\infty$$
which implies $\limsup_{t \to \infty} \frac{B_t ^2}{t \ln t} \leq 1$ (otherwise, there exists $\delta > 0$, such that $\frac{B_t ^2}{t \ln t} >1+\delta$ infinitely often as $t \to \infty$, so $ \frac{B_t^2}{t \ln t}- \frac{(1+2t)\ln(1+2t)}{2t \ln t} >0$ infinitely often for large $t$ with positive probability). Note that $$\frac{\ln(1+2t)}{\ln t} \to 1$$ using L'Hospital's rule.
