Consider a finite set S of positive integers, and define $q(S) = \sum_{s \in S}{1/s^2}$. Letting $\rho = \pi^2/6$, we have $q(S)$ in the ranges $[0, \rho - 1), [1, \rho)$. I conjecture that for every rational $r$ within those ranges, $\exists S: q(S) = r$. Can anyone prove or disprove this?
Note that for plain unit fractions $u(S) = \sum_{s \in S}{1/s}$ it is straightforward to show by a greedy algorithm that $\exists S: u(S) = r \forall r \in \mathbb{Q}^+$; conversely for prime unit fractions $p(S) = \sum_{s \in S}{1/p_s}$ almost all rationals are unreachable, since for each $S$, $p(S)$ has a unique (squarefree) denominator. Given that the primes are denser than the squares, it would thus be a somewhat interesting result if the conjecture were to be proven.
Computationally I have established existence of $S$ for all rationals with denominators up to 50 within the range; see results on github.
Update: added "finite" to clarify first sentence; corrected $\mathbb{R}$ to $\mathbb{Q}$.
You have rediscovered a celebrated theorem of R. L. Graham, who proved this exact statement (as a special case of something much more general) in 1964.