random 2-dimensional vectors, $X_n$ is uniformly distributed Let $(X_1 ,X_2, \ldots)$ be an independent sequence of random 2-dimensional vectors, where for each $n$, $X_n$ is uniformly distributed on the squares with vertices  $[\pm n,\pm n]$. 
How can I calculate the probability $$ P(\{w: |X_n(w)| \rightarrow \infty \ \ as \ n \rightarrow \infty\}).$$
My attempt: 
Let $A_n=\{ w \in \Omega | |X_k| \geq k\}$ with $k \in \mathbb{Z}^{+}$, then $$ P(A_n)= \left\{ \begin{array}{lcc}
             0 &   if  & k \geq \sqrt{2}n \\
             \\ 1-\frac{k^2 \pi}{n^2} &  if & k \leq n \\
             \\ c(k) &  if  & n < k < \sqrt{2}n 
             \end{array}
   \right. $$
for $c(k)$ a certain value depending of $k$.
I that correct? Or How can I compute the probability on this $A_n=\{|X_n(w)| < n^\alpha\}$ ?
Could someone help pls? Thanks for your time and help.
 A: Why not use the same approach I showed you in your previous question ?
Since all norms on $\mathbb R^2$ are equivalent, let $\|\cdot\|$ denote the maximum norm (its unit ball is a square, how convenient...).
Note that $\displaystyle P(\|X_n\|\to \infty)= P\left(\bigcap_N\bigcup_n \bigcap_{k\geq n}(\|X_k\| \geq N)\right)$.
Since the sequence of sets (indexed by $N$) $\bigcup_n \bigcap_{k\geq n}(\|X_k\| \geq N)$ is decreasing, $$P\left(\bigcap_N\bigcup_n \bigcap_{k\geq n}(\|X_k\| \geq N)\right)=\lim_N P\left(\bigcup_n \bigcap_{k\geq n}(\|X_k\| \geq N)\right)$$
so it suffices to compute $P\left(\bigcup_n \bigcap_{k\geq n}(\|X_k\| \geq N)\right)$ for each $N$.
But $P\left(\bigcup_n \bigcap_{k\geq n}(\|X_k\| \geq N)\right)$ is nothing other than $$P(\liminf_n \|X_n\|\geq N) = 1-P(\limsup_n \|X_n\|<N)$$
Since $\sum_{n=1}^\infty P(\|X_n\|< N)=\sum_{n=1}^N 1 + \sum_{n=N+1}^\infty \frac{N^2}{n^2}< \infty$, Borel-Cantelli lemma yields $P(\limsup_n \|X_n\|<N) = 0$, hence $P(\liminf_n \|X_n\|\geq N)=1$ for all $N$.
Therefore, $P(\|X_n\|\to \infty)=1$.
A: If $X_n\not\to\infty$, then some subsequence must remain bounded, which for large $n$ is unlikely. To make this idea a little more formal, take a sequence $(a_n)$ that goes to infinity slowly ($a_n=\log n$ will do) and show that $P(|X_n|\le a_n) $ is summable. You can then apply Borel-Cantelli. 
