Compute
\[\sum_{1 \le j < i} \frac{1}{2^{i + j}},\]where the sum is taken over all positive integers $i$ and $j$ such that $1 \le j < i.$
We have that
\begin{align*}
\sum_{1 \le j < i} \frac{1}{2^{i + j}} &= \sum_{j = 1}^\infty \sum_{i = j + 1}^\infty \frac{1}{2^{i + j}} \\
&= \sum_{j = 1}^\infty \frac{1}{2^j} \sum_{i = j + 1}^\infty \frac{1}{2^i} \\
&= \sum_{j = 1}^\infty \frac{1}{2^j} \left( \frac{1}{2^{j + 1}} + \frac{1}{2^{j + 2}} + \frac{1}{2^{j + 3}} + \dotsb \right) \\
&= \sum_{j = 1}^\infty \frac{1}{2^j} \cdot \frac{1/2^{j + 1}}{1 - 1/2} \\
&= \sum_{j = 1}^\infty \frac{1}{2^j} \cdot \frac{1}{2^j} \\
&= \sum_{j = 1}^\infty \frac{1}{4^j} \\
&= \frac{1/4}{1 - 1/4} \\
&= \boxed{\frac{1}{3}}.
\end{align*}