Variance of model with Bernoulli-distributed random variables Let $(Y_n)$ and $(Z_n)$ be two independent sequences of iid Bernoulli-distributed random variables with parameter $p$ and $q$, respectively. For $n=1,2,3,...$, define $$ X_n = 3Y_{n+1} + 4Y_{n+1}Z_{n-3} + Y_{n+3}Z_{n-1}$$
Question: Show that the variance of $S_n = \sum_{i=1}^n X_i $ is converging for $n$ going to infinity, i.e. $ \lim_{n \rightarrow \infty} var(S_n) < \infty$.
My "Ansatz":
First, I observed that $X_3$ is dependent $X_1$ and that $X_5$ is dependent on $X_3$, and so on. The same behaviour for $X_n$ for $n$ being even, i.e. $X_4$ being dependent on $X_2$. So I can decompose $S_n$ into $S_n^* = \sum_{i=2,4,6,...}X_i$ and $S_n^{**} = \sum_{i=1,3,5,...}X_i$. For simplicity assume that $n$ is some number number which can be divided by 4 (e.g. 48), s.t. $S_n^*$ and $S_n^{**}$ are of the same size. Then one can state that $$ var(S_n) = var(S_n^{*}+S_n^{**}) = 2*var(S_n^{*})$$ 
 A: I'm not sure how $X_2$ is defined unless we index the $Z_i$ with negative indices, so I started the sum from $i=4$. Also, the variance is clearly unbounded for 
$$ \lim_{n\to\infty}\sum_{i=4}^n X_i, $$
so I'm assuming instead you mean the standard normalization
$$ \lim_{n\to\infty}\frac{1}{\sqrt{n}}\sum_{i=4}^nX_i. $$
I'm going to compute what you called $\text{Var}(S^*_n)$, but with the $1/\sqrt{n}$ normalization above.
Writing out the first few terms of the sequence, it's easy to see that, for even $n\geqslant6$, 
$$ \tag{1}\label{eqn:1}\sum_{i=4,6,\dots}^{n}X_i=Y_5(3+4Z_1)+Y_{n+3}Z_{n-1}+\sum_{i=6,8,\dots,}^{n}Y_{i+1}(3+5Z_{i-3}) $$
Since each summand in the sum above is independent, in light of (\ref{eqn:1}), we have that
\begin{align}
\text{Var}\bigg(\sum_{i=4,6,\dots}^nX_i\bigg)
&=\text{Var}\bigg(Y_5(3+4Z_1)+Y_{n+3}Z_{n-1}+\sum_{i=6,8,\dots,}^{n}Y_{i+1}(3+5Z_{i-3})\bigg)\\
&=\underbrace{\text{Var}(Y_5(3+4Z_1))}_{\displaystyle\equiv\alpha}+\underbrace{\text{Var}(Y_{n+3}Z_{n-1})}_{\displaystyle\equiv\beta}+\sum_{i=6,8,\dots,}^{n}\underbrace{\text{Var}(Y_{i+1}(3+5Z_{i-3}))}_{\displaystyle\equiv\gamma}\\
&=\alpha+\beta+\big(\frac{n-4}{2}\big)\gamma
\end{align}
Note that the constants $\alpha,\beta$ and $\gamma$ are independent of $n$. Our result is then
\begin{align}
\text{Var}\bigg(\frac{1}{\sqrt{n}}\sum_{i=4,6,\dots}^nX_i\bigg)
&=\frac{1}{n}\bigg(\alpha+\beta+\big(\frac{n-4}{2}\big)\gamma\bigg)\\
&=\frac{\alpha+\beta}{n}+\bigg(\frac{1}{2}-\frac{2}{n}\bigg)\gamma
\end{align}
which is clearly finite as $n\to\infty$. My calculations give that
$$ \gamma=25q(1-q)p+p(1-p)(3+5q)^2. $$
If I made any algebraic mistakes let me know!
