Expectation and Variance of Infinite Sum

Question

I am presented with the random variable $S = \sum_{k=1}^{\infty} a^{-k} Z_k$ where $a > 1$ and $Z_k$ is defined as $P(Z_k = \pm 1) = 0.5$. As part of an $L^2$ convergence proof, I am tasked with proving $S \in L^2$ as well as finding $\text{Var}(S)$. Now, proving $S \in L^2$ seems simple enough: \begin{align*} \text{E}\left[ \vert S \vert^2 \right] &= \text{E}\left[ \left\vert \sum_{k=1}^{\infty} a^{-k} Z_k \right\vert^2 \right]\\ &\leq \text{E}\left[ \left(\sum_{k=1}^{\infty} a^{-k} \left\vert Z_k \right\vert \right)^2 \right]\\ &= \text{E}\left[ \left(\sum_{k=1}^{\infty} a^{-k} \right)^2 \right]\\ &= \text{E}\left[ \left( \frac{1}{a-1} \right)^2 \right]\\ &= \left( \frac{1}{a-1} \right)^2 \end{align*} Thus, $S \in L^2$ since $\text{E}\left[ \vert S \vert^2 \right]^{\frac{1}{2}} < \infty$. However, I am struggling to determine the expression for $\text{Var}(S)$, or more specifically what $\text{E}\left[ S^2 \right]$ is equal to since it is clear (by the Lebesgue Dominated Convergence Theorem) that $\text{E}[S] = 0$. The provided final answer is that $$ \text{Var}(S) = \frac{1}{a^2 - 1} $$ which makes sense by the (linearity and coefficient) properties of variance $$ \text{Var}(S) = \text{Var}\left( \sum_{k=1}^{\infty} a^{-k} Z_k \right) = \sum_{k=1}^{\infty} (a^{-2})^k \text{Var}(Z_k) = \sum_{k=1}^{\infty} (a^{-2})^k = \frac{1}{a^2 - 1} $$ but it is not immediately obvious to me how $$ \text{E}\left[ S^2 \right] = \text{E}\left[ \left( \sum_{k=1}^{\infty} a^{-k} Z_k \right)^2 \right] = \text{E}\left[ \sum_{k=1}^{\infty} a^{-2k} Z_k^2 \right] = \sum_{k=1}^{\infty} (a^{-2})^k = \frac{1}{a^2 - 1} $$ Typically, I would simply use the properties of variance to get the provided final result, but I have been instructed that this is "improper," so is there a way to prove $$ \text{E}\left[ \left( \sum_{k=1}^{\infty} a^{-k} Z_k \right)^2 \right] = \text{E}\left[ \sum_{k=1}^{\infty} \left( a^{-k} Z_k \right)^2 \right] $$ or perhaps I am missing something quite simple? Any help/hints/guidance is well appreciated.

Answer 1

$ E[Z_i Z_j] = E[Z_i] E[Z_j] = 0 $ for $i \neq j$