quadratic variation of brownian motion doesn't converge almost surely I just came across the following remark:
If $(B_t)_{t\geq0}$ is a one dimensional Brownian motion and if we have a subdivison $0=t_0^n<...<t_{k_n}^n=t$ such that $\sup_{1\leq i\leq k_n}(t_i^n-t_{i-1}^n)$ converges to $0$ is $n$ converges to $\infty$ then $\lim_{n\to\infty}\sum_{i=1}^{k_n}(B_{t_i^n}-B_{t_{i-1}^n})^2=t$ in $L^2$. If $\sum_{n\geq 1}\sum_{i=1}^{k_n} (t_i^n-t_{i-1}^n)^2<\infty$, then we also have almost sure convergence
Now I'm trying to find a subdivison so that $\lim_{n\to\infty}\sum_{i=1}^{k_n}(B_{t_i^n}-B_{t_{i-1}^n})^2$ does not converge almost surely and I'm kinda stucked. 
My idea was the following: First note that the subdivision needs to fulfill the following: $\sum_{n\geq 1}\sum_{i=1}^{k_n} (t_i^n-t_{i-1}^n)^2=\infty$. Next I thought about using Borel-Cantelli, which says: If $\sum_{n=1}^{\infty} \mathbb{P}(E_n)=\infty$ and if the events $E_n$ are independent, we have that $\mathbb{P}(\limsup E_n=\infty)=1$. 
Now if we can create/formulate events $E_n$ s.t. $\mathbb{P}(E_n)=\sum_{i=1}^{k_n}(t_i^n-t_{i-1}^n)^2$ and such that $\limsup E_n=lim_{n\to\infty}\sum_{i=1}^{k_n} (B_{t_i^n}-B_{t_{i-1}^n})^2$, then Borel-Cantelli would tell us that $\sum_{i=1}^{k_n} (B_{t_i^n}-B_{t_{i-1}^n})^2$ would not converge almost surely. 
My problem is that I can't see how in this case $E_n$ needs to be defined. (And I'm not sure whether this is a good approach...)Can anyone help me? Thanks in advance!

Consider the sequence of partitions $$\pi(n) = \bigcup_{i=0}^n \left\{\frac in t \right\},\ n\geqslant1 $$ of $[0,t]$, that is, $t_i^n = \frac int$ for $0\leqslant i\leqslant n$. Then $$\sup_{1\leqslant i\leqslant n}\left(t_i^n-t_{i-1}^n\right) = \frac1n\stackrel{n\to\infty}\longrightarrow0, $$ but $$\sum_{n=1}^\infty\frac1n=\infty. $$
From here we may use the second Borel-Cantelli lemma to show that $\sum_{i=1}^n \left(B_{\frac itn}-B_{\frac{i-1}tn}\right)^2$ does not converge to $t$ almost surely. However, this is somewhat of a moot point, considering that $L^2$ convergence implies convergence in probability, which in turn implies a.s. convergence along a subsequence.
