Show that this sequence does not obey the SLLN The exercise reads:
Let $(\sigma_n)$ be a sequence of real numbers for which
$$ \sum_{n=1}^{\infty} \frac{1}{n^2} \sigma^2_n=\infty$$ Prove that there exist an independent sequence $(X_n)$ of integrable real random variables which statisfy $V(X_n)=\sigma^2_n$ for every $n$, but which does not obey the strong law of large numbers (SLLN).
Hints: Set $\alpha_n:=\max\{\sigma_n,n\}$, $\beta_n:=\min\{\sigma_n,n\}$ and define $X_n$ such that
$$ P\{X_n=\alpha_n\}=P\{X_n=-\alpha_n\}=\frac{1}{2}\bigg(\frac{\beta_n}{n}\bigg)^2,$$
$$ P\{X_n=0\}=1-\frac{1}{2}\bigg(\frac{\beta_n}{n}\bigg)^2$$
Then use Exercise 4 which says that if $(X_n)$ is a independent sequence of centered integrable real random variables that obeys the SLLN then
$$\sum_{n=1}^{\infty} P\{n^{-1}|X_n|\geq \epsilon \} < \infty \hspace{0.5cm} \text{for every } \epsilon>0$$
Here's what I did:
Let $(X_n)$ be an independent sequence of real random variables distributed as above (such a sequence exists by the theory of infinite products of probability spaces). Then we see that $E(X_n)=0$ and $V(X_n)=\sigma^2_n$ for each $n$ as desired. Suppose on the contrary that this sequence obeys the SLLN. Then Exercise 4 with $\epsilon=1$ implies that
$$\sum_{n=1}^{\infty} P\{n^{-1}|X_n|\geq 1 \} < \infty$$
Since $\frac{\alpha_n}{n}\geq 1$ for each $n$ we have in particular
$$\sum_{n=1}^{\infty} P\bigg \{n^{-1}|X_n|  \geq \frac{\alpha_n}{n} \bigg \} =\sum_{n=1}^{\infty}  \bigg(\frac{\beta_n}{n}\bigg)^2 <\infty$$
But because $\big(\frac{\beta_n}{n}\big)^2\leq \frac{\sigma^2_n}{n^2}$ I don't see how I can obtain a contradiction.
Any ideas?

The series $\sum_{n=1}^{\infty}  \bigg(\frac{\beta_n}{n}\bigg)^2 $ is divergent:
indeed,
$$
\frac{\beta_n^2}{n^2}=\min\left\{\frac{\sigma_n^2}{n^2},1\right\};
$$
if $\sigma_n/n\to 0$, then $\beta_n^2/n^2=\sigma_n^2/n^2$ for $n$ large enough; if $\sigma/n$ does not converge to $0$, then $\beta_n^2/n^2$ does not convergence either to $0$ hence there is also divergence.
