The decimal representation of $\frac{1}{n}$ terminates if and only if $n=2^{i} 5^{j}$ for some nonnegative integers $i, j$, so our desired sum is $$\sum_{i \geq 0} \sum_{j \geq 0} 2^{-i} 5^{-j}=\sum_{i \geq 0} 2^{-i} \sum_{j \geq 0} 5^{-j}=\left(1-2^{-1}\right)^{-1}\left(1-5^{-1}\right)^{-1}=\frac{2}{1} \frac{5}{4}=\frac{5}{2}$$