How to show convergence a.s. when sum of $P(A_n)$ is $\infty$ and the sequence is not independent

There is a sequence of random variables defined by $$Y_n = \Big(\Big|{1-\frac\Theta \pi}\Big|\Big)^n$$ where $$\Theta\sim\mathrm{unif}[0,2\pi].$$ I have shown that the sequence converges to $$0$$ in probability. Applying Borel Cantelli Lemma by taking $$A_n = \{|Y_n|>0\}$$ and then evaluating $$\sum_{n=1}^\infty P(A_n)$$, I get it to be $$\infty$$. Since the random variables are not independent, I can't conclude anything from this result. How do I prove/disprove convergence almost surely?

$$Z=(1-\Theta/\pi)$$ is almost surely in $$(-1,1),$$ in which case $$Z^n\to 0.$$ Thus $$Y_n = Z^n \to 0$$ almost surely.
• Thank you for you answer. $P(\lim_{n\to\infty}Y_n = 0) = 1$ implies a.s. convergence right? In this case $Y_n \to 0$ a.s. as $n \to \infty$. Does that guarantee a.s. convergence? – learner Nov 3 '18 at 6:34
• @learner If it almost surely converges to zero, then it almost surely converges. More formally, $P(Y_n \mbox{ converges}) \ge P(Y_n\to 0) = P(Z^n\to 0) \ge P(|Z|<1) = 1.$ – spaceisdarkgreen Nov 3 '18 at 6:41