I have the following problem:
Let $X_n$, $n \geq 1$ be independent r.v. identically distributed with the probability function:
$P(X_{i} = k) = P(X_{i} = - k) = \frac{p}{2} q^{k-1}$, $k = 1,2,...,m$, $P(X_{i} = m + 1) = P(X_{i} = -m -1) = \frac{q^{m}}{2}$
where $m>1$, $p\in (0,1)$ and $q = 1-p$. Calculate the almost surely limit of $Z_{n} = \frac{\frac{X_{1}}{X_{2}} + \frac{X_{3}}{X_{4}} +...+ \frac{X_{2n-1}}{X_{2n}}}{X_{1}^{2} + X_{2}^{2} + ... + X_{n}^{2}}$.
The exercise also adds a suggestion that is as follows: Prove that the numerator's limit is $0$ and that the denominator's limit is always greater than $0$.
So far I think I can say that the denominator's limit must be greater than $0$ because for all $x\in \mathbb{R}-\{0\}$, $x^{2} > 0$. I know that $X_{i} \neq 0$ and, therefore, $X_{1}^{2} + X_{2}^{2} + ... + X_{n}^{2} > 0$.
Let me know if there's any mistake in my reasoning and also if you find any way to proceed to find the answer to the problem.
I'll give you another hint. Note that we may write $$Z_n = \frac{W_n}{Y_n}$$ where $W_n = \frac{1}{n}\left( \frac{X_1}{X_2} + \frac{X_3}{X_4} + \ldots + \frac{X_{2n-1}}{2n} \right)$ and $Y_n = \frac{1}{n}\left( X_1^2 + X_2^2 + \ldots +X_n^2 \right)$. Notice that both $W_n$ and $Y_n$ are the means of iid random variables with finite mean. The strong law of large numbers tells us that $$W_n \xrightarrow{a.s.} E\left[ \frac{X_1}{X_2} \right] $$ and $$Y_n \xrightarrow{a.s.} E[X_1^2]>0$$ Can you compute these expectations and take it from there?