HMMT-Nov 2016 guts, Problem 19

Let $S$ be the set of all positive integers whose prime factorizations only contain powers of the primes 2 and 2017 (1, powers of 2, and powers of 2017 are thus contained in $S$ ). Compute $\sum_{s \in S} \frac{1}{s}$.

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

Answer: \frac{2017}{1008}