I'm preparing for final test at University doing a bank of exercises and I found this one and I don't know how to even start it: $$$$ Let $ X_{1},X_{2},X_{3}... $ be independent variables with the same distribution of value expected $\mu$ and finite variance. Let $ Y_{n} $ are variables with the Pareto distribution $ P(Y_{n}>u)=(1+u)^{-n} $. Examine almost sure convergence of series $ Z_{n} $, $ n \ge 1 $, where $$ Z_{n} = \left| \frac{ \sum_{1 \le i<j \le n} X_{i} X_{j}}{ {n \choose 2} } \right|^{1+ Y_{n} } $$ $$$$ $$$$
I'm also very sorry for every mistake at translation it from my language because in my country there is no math forum where I can find the solution of this problem.
Firstly, we'll show that $Y_n$ converges to $0$, which is a straightforward application of Borell-Cantelli. Take any $m \in \mathbb N_+$ and note that $\sum_{n=1}^\infty \mathbb P(Y_n > \frac{1}{m}) = \sum_{n=1}^\infty (1+\frac{1}{m})^{-n} < \infty$, since it is a geometric series. By Borell-Cantelli, we know that there exists $\Omega_m$: $\mathbb P(\Omega_m)=1$ and for any $\omega \in \Omega_n$ $\exists_{N(\omega) \in \mathbb N_+} \forall_{n > N(\omega)}: Y_n(\omega) < \frac{1}{m}$. Now let $\Omega^* = \bigcap_{m=1}^\infty \Omega_m$. It is of $1-$measure, as a countable intersection of $1$-measure sets. Note that for any $\omega \in \Omega_m$ we have $Y_n(\omega) \to 0$ (since for any $\varepsilon > 0$ we have $m \in \mathbb N_+$ such that $\frac{1}{m}<\varepsilon$ and use the fact that $\Omega_m \subset \Omega^*$ to conclude). It means that $Y_n \to 0$ almost surely.
As for the second part, it is again, a straightforward application of SLLN. Note that (adding and substracting $\sum_{j=1}^n X_j^2$) $$ 2\sum_{1 \le i < j \le n} X_iX_j = \sum_{i=1}^n \sum_{j=1}^n X_iX_j - \sum_{j=1}^nX_j^2 = (\sum_{j=1}^n X_j)^2 - \sum_{j=1}^n X_j^2$$ Hence $$ \left| \frac{2\sum_{1 \le i < j \le n}X_iX_j}{n(n-1)} \right|= \frac{n^2}{n(n-1)}\left| (\frac{\sum_{j=1}^n X_j}{n})^2 - \frac{1}{n}\frac{\sum_{j=1}^n X_j^2}{n}\right|$$ As we know, $\frac{1}{n}\sum_{j=1}^n X_j \to \mu$ almost surely and $\frac{1}{n} \sum_{j=1}^n X_j^2 \to \mathbb E[X_1^2]$ almost surely (by assumption, we know it is finite). By intersecting those $2$ sets of full measure, we finally gets that Z$_n$ converges almost surely, to:
$$ Z_n = \left |\frac{2 \sum_{1 \le i < j \le n}X_iX_j}{n(n-1)}\right|^{1+Y_n} \to \Big(1 \cdot \left| \mu^2 - 0 \cdot \mathbb E[X_1^2] \right| \Big)^{1+0} = \mu^2 $$