= Solution
The $i$th screen distance is $x_0/2^i$, starting at $i=1$. The previous <expected value> therefore scales to $E[Y_i]=x_0\ln3/(\pi2^i)$. To justify the infinite <linearity of expectation>, note that $|\tan\theta_i|\le\sqrt3$ throughout the restricted angular interval. Thus
$$
\sum_{i=1}^\infty|Y_i|\le\sqrt3x_0\sum_{i=1}^\infty2^{-i}=\sqrt3x_0.
$$
This deterministic integrable bound justifies the <expectation of an absolutely summable random series>, even without assuming independent firing angles. Summing the <geometric series> gives
$$
\boxed{E\left[\sum_{i=1}^\infty Y_i\right]=\frac{x_0\ln3}{\pi}\sum_{i=1}^\infty2^{-i}=\frac{x_0\ln3}{\pi}.}
$$
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