limes of martingale given quadratic variation 
Given a martingale $M_t$ with $M_0=0$ and $E[M_t^2]<\infty$. Let the
  quadratic variation of $M_t$ be $[M]_t:=ln(1+t)$. Calculate $\limsup_{t\to\infty} \dfrac{M_t}{\sqrt{t}}$.

Does anyone have an approach for that? I thought of using Itô's formula but this didn't help.

Let $\epsilon>0$, we define $$A_n=\{\max_{2^n\leq t\leq 2^{n+1}}\frac{|M_t|}{\sqrt{t}}>\epsilon\}$$
We have
 $$A_n\subseteq{\{\exists t \in[2^{n},2^{n+1}] , |M_t|>\epsilon\sqrt{2^n} \}}$$,
Thus,
 $$P(A_n) \leq P({\{\exists t \in[2^{n},2^{n+1}] , |M_t|>\epsilon\sqrt{2^n} \}})$$
or equivalently ,
 $$P(A_n) \leq P({\max_{2^n\leq t\leq 2^{n+1}} |M_t|>\epsilon\sqrt{2^n} })$$
A that stage, you can apply the Doob inequality  : 
$$ P({\max_{2^n\leq t\leq 2^{n+1}} |M_t|>\epsilon\sqrt{2^n} })\leq\frac{E(M_{2^{n+1}}^2)}{\epsilon^22^n}$$
We have that $$M_t^2=U_t+log(1+t)$$ where $U_t$ is a martingale,
therefore 
$$ P(A_n)\leq\frac{log(1+2^{n+1})}{\epsilon^22^n}$$
The Borel-Cantelli lemma concludes that 
$$P(\limsup_{t \to \infty}\frac{|M_t|}{\sqrt{t}} \leq \epsilon)=1 $$
Moreover, 
$$\{\frac{|M_t|}{\sqrt{t}}=0\}=\cap_{k\geq1}\{\limsup_{t \to \infty}\frac{|M_t|}{\sqrt{t}}\leq\frac{1}{k}\}$$
You can conclude
