Almost sure convergence of series

Question

I've been trying to prove that the series $ \sum_{i=1}^n X_iY_i $ converges almost surely. I know that $X_i$ is $Bern(\frac{1}{i^2})$, that $E(Y_i)=0$ and that $E(Y_i^2)=i^2$. Furthermore, all $X_i$ and $Y_i$ are independent. I've shown that $$ \sum_{i=1}^\infty P(|X_i Y_i|>\varepsilon)<\infty \qquad \sum_{i=1}^\infty E(X_iY_i\mathrm{1}_{(|X_iY_i|\leq\varepsilon)})<\infty $$ however the last sum, that is $\sum_{i=1}^\infty V(X_iY_i\mathrm{1}_{(|X_iY_i|\leq\varepsilon)})$ in Kolmogorov's three series theorem I cannot crack. Nor can I use Khinchin-Kolmogorovs convergence theorem, since $V(X_iY_i)=1$, so the sum of those alone is clearly divergent. I am now in need of a hint on how to show that the series converges. Any help will be much appreciated.

Answer 1

For $\epsilon>0$, \begin{align} \mathsf{E}[(X_iY_i)^2 1\{|X_iY_i|\le \epsilon\}]&=\mathsf{E}[(X_iY_i)^2 1\{|X_iY_i|\le \epsilon\}1\{X_i=0\}] \\ &\quad+\mathsf{E}[(X_iY_i)^2 1\{|X_iY_i|\le \epsilon\}1\{X_i=1\}] \\ &=\mathsf{E}[Y_i^2 1\{|Y_i|\le \epsilon\}1\{X_i=1\}] \\ &=\mathsf{E}[Y_i^2 1\{|Y_i|\le \epsilon\}]\,\mathsf{P}(X_i=1)\le (\epsilon/i)^2. \end{align} Thus, $$ \sum_{i\ge 1}\operatorname{Var}(X_iY_i 1\{|X_iY_i|\le \epsilon\})\le \sum_{i\ge 1}(\epsilon/i)^2<\infty. $$