Where is $|\theta|<1$ used in recursive method derivation of invertibility of MA(1)? Silly question: From what I understand, an MA process is invertible when it can be represented as an AR($\infty$) process.
When using lag operator, it is somewhat clear that $|\theta|<1$ is required. But I am unable to see this in following derivation.
Consider the following MA(1) process:
\begin{align}
x_t &= \epsilon_t + \theta \epsilon_{t-1} \\
&= \epsilon_t + \theta (x_{t -1}- \theta \epsilon_{t-2}) \\
&= \epsilon_t + \theta x_{t -1}- \theta^2 (x_t- \theta\epsilon_{t-3})\\
&=\epsilon_t - \sum\limits_{j=1}^{\infty}(-\theta)^jx_{t-j}
\end{align}
and so on.
So we are able to express MA(1) process as an AR($\infty$).
At which step am I using $|\theta|<1$?

This is actually somewhat subtle. When $|\theta|>1$, you can still express $\epsilon_t$ as AR($\infty$) on $x_t, x_{t+1}, x_{t+2}, ...$.  But $\epsilon_t$ is no longer the fundamental innovation on $x_t$, defined as $x_t - E(x_t|x_{t-1}, x_{t-2}, ...)$.
So if being "invertible" means a MA process on the "fundamental" innovations, then root conditions like $|\theta|<1$ is needed.
