Let $\xi_n \sim N(0,1)$ and $A_n = \frac{\sqrt 2}{\pi n} \xi _n$.
Show that $\sum_{i=1}^\infty |A_i| = \infty$ $\mathbb P$-a.s.
MY ATTEMPT:
Usually when trying to prove or refute almost sure convergence or divergence of a sequence, Borel-Cantelli lemma can be handy (however, this approach can be completely off the mark).
I want to prove:
$$\forall N \in \mathbb N \quad \exists m \in \mathbb N \quad \forall k \geq m \quad \sum _{i=1}^k |A_i| > N \quad \mathbb P -\text{a.s.}$$
$$ \begin{align*} \Leftrightarrow &\quad \mathbb P \left(\bigcap_{N\in \mathbb N}\bigcup _{m \in \mathbb N} \bigcap _{k \geq m} \left\{\sum_{i=1}^k|A_i| > N\right\}\right) = 1\\ \Leftrightarrow &\quad \mathbb P \left(\bigcup_{N\in \mathbb N}\bigcap _{m \in \mathbb N} \bigcup _{k \geq m} \left\{\sum_{i=1}^k|A_i| \leq N \right\}\right) = 0\\ \Leftrightarrow &\quad \mathbb P \left(\bigcup_{N\in \mathbb N} \limsup_{m \rightarrow \infty} B_{m, N}\right) = 0 \end{align*} $$
$$ \text{with} \quad B_{k, N} := \left\{\sum_{i=1}^k|A_i| \leq N \right\}. $$
Now if we could show
$$ \forall N \in \mathbb N \quad \sum _{m=1} ^ \infty \mathbb P \left( B_{m, N}\right) = \sum _{m=1} ^ \infty \mathbb P \left( \sum_{i=1}^m|A_i| \leq N \right)< \infty $$
then with subadditivity of $\mathbb P$ and Borel-Cantelli we would get
$$\mathbb P \left(\bigcup_{N\in \mathbb N} \limsup_{m \rightarrow \infty} B_{m, N}\right) \leq \sum_{N=1}^\infty \mathbb P \left(\limsup_{m \rightarrow \infty} B_{m, N}\right) = 0.$$
I know that $\sum_{i=1}^\infty |A_i|$ is the sum of folded-normal distributed random variables. However, I can't find a way to control the quantity $$\sum _{m=1} ^ \infty \mathbb P \left( \sum_{i=1}^m|A_i| \leq N \right).$$
A rather creative approach was presented in this answer, however, in my case summands are not standard normal (but rather rescaled with $\frac {\sqrt 2 }{\pi n}$), so I can't use the symmetry of multivariate normal as suggested.
Thanks in advance!
