Given $X_1, ...$ defined as above for some sequence of $a_n>0$, and let $Y_n = \prod_{i=1}^n X_i$, show $Y_n \stackrel{a.s}{\rightarrow} 0$ if $\limsup_{n \rightarrow \infty} a_n < 2$.
Here's my attempt, will be grateful if someone helps me to check this.
To show $Y_n \stackrel{a.s}{\rightarrow} 0$, it suffices to show that $\mathbb{P}(|X_n| \geq 1 \, \text{i.o.}) = 0$. To show this, assume the extreme case that $a_1, a_2, ... > 1$, and compute $$ \sum_{i=1}^\infty \mathbb{E} \left[1_{\{|X_i|<1\}} - 1_{\{|X_i|> 1\}}\right] = \sum_{i=1}^\infty \frac{2}{2a_i} - \frac{2(a_i-1)}{2a_i} = \sum_{i=1}^n \frac{1}{a_i} - \frac{1}{2} = \infty, $$ since $\limsup a_n < 2$. Therefore, $|X_i| < 1$ infinitely often. Hence, $Y_n = \prod_{i=1}^n X_i = 0$ almost surely.
My second question is, will the a.s. convergence of $Y_n$ still hold if its $\limsup a_n \leq 2$? I suspect not, as $\limsup a_n = 2$ could just mean a constant sequence $a_1 = a_2 = ... = 2$. But how can I prove this claim?