Let $\left(\theta_{n}\right)_{n\in\mathbb{N}}$ be a sequence of random variables iid with uniform distribution in $\left\{-1,1\right\}$. We consider $X_{n}$ recurrent random sequence given by $$X_{n}=a X_{n-1}+\theta_{n}$$ where $a\in (1,\infty)$.
Show that $$\mathrm{P}\left( \lim_{n\rightarrow \infty} X_{n} \in \left\{-\infty,\infty \right\} \right)=1.$$
Remark: I need a suggestion to prove this fact.
Define a random variable $Y = \sum_{k=0}^{\infty} a^{-k}\theta_k$. Note that this sum converges almost surely since $|\theta_k|<1$ for all $k$, and because $\sum_k a^{-k}=a/(a-1)<\infty$.
Next, notice by induction that $X_n = a^n \theta_0+ a^{n-1} \theta_1 + ... + a \theta_{n-1} + \theta_n$. Therefore $a^{-n}X_n = \sum_{k=0}^n a^{-k}\theta_k$.
So we see that $|a^{-n}X_n-Y| \leq \sum_{k=n+1}^{\infty} a^{-k}|\theta_k| \leq \sum_{k=n+1}^{\infty} a^{-k}= a^{-n}/(1-a)$. Multiplying both sides by $a^n$, we see that $|X_n-a^nY| \leq 1/(1-a)=:C$ almost surely.
Hence, on the event $\{Y>0\}$, we see that $X_n>a^nY - C$ which diverges to $+\infty$ as $n \to \infty$ (since $Y>0$ and $a^n \to \infty$). Similarly, on the event $\{Y<0\}$, we see that $X_n < C - a^n(-Y)\to -\infty$. Now, it is not difficult to see that $Y$ has no atoms, so that $P(Y=0)=0$. So we are finished.