Solution:
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
$$\left(\frac{1}{1-\frac{1}{2}}\right) \cdot \frac{1}{1-\frac{1}{2017}} = 2 \cdot \frac{2017}{2016} = \frac{2017}{1008}.$$