Almost sure and $\mathcal{L}^1$ convergence of $Y_n=(\cos a)^{-n}\cos(a(Z_1+\cdots+Z_n))$ with $(Z_n)$ i.i.d. Bernoulli

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Let $(X_i)$ be i.i.d. with $P(X_i= a)=P(X_i = -a)=\frac{1}{2}$, for some $a$ such that $2a \notin\pi\mathbb Z$. Let $$Y_n=\frac1{\cos^n(a)}\cos\left(\sum_{i=1}^n X_i\right).$$ Check whether $(Y_n)$ converges almost surely and in $\mathcal{L}^1$.

I usually don't have problems proving that a sequence of random variables converges. The intuition here is that the denominator is a number in $(0,1)$ raised to infinity, so it converges to $0$, while the numerator is bounded - but that is only the intuition and does not bring me any closer to proving that this does not converge almost surely and in $\mathcal{L}^1$. I also proved that $(Y_n)$ is a martingale but that does not yield me any useful information.

Would anyone like to share their thoughts on this task?

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This is not a complete answer, just some ideas too long to fit into a comment.

Since only cosines are involved, you can understand $\sum_{i=1}^n X_i$ as a random walk on a circle. Then it is not hard to see that your random walk visits any neighborhood of $0$ infinitely often. So the $\limsup$ of numerator is $1$, while the denominator converges to $0$, hence the almost sure convergence is impossible.

If it were converging in $L^1$, then thanks to the martingale property and the martingale convergence theorem, it would converge almost surely as well. Therefore, there is no $L^1$-convergence either.