Since every $s$ can be written as $2^{i} \cdot 2017^{j}$ for non-negative integers $i$ and $j$, the given sum can be written as $\left(\sum_{i=0}^{\infty} \frac{1}{2^{i}}\right)\left(\sum_{j=0}^{\infty} \frac{1}{2017^{j}}\right)$. We can easily find the sum of these geometric series since they both have common ratio of magnitude less than 1, giving us $\left.\left(\frac{1}{1-\frac{1}{2}}\right) \cdot \frac{1}{1-\frac{1}{2017}}\right)=\frac{2}{1} \cdot \frac{2017}{2016}=\frac{2017}{1008}$.