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You are likely misunderstanding what the author writes in the proof of Hurwitz's Theorem. Specifically, he says, "if $S$ a finite set of rational numbers, then we can take $n$ large enough...", etc. This means that for the given set of rationals $S$ (not necessarily a set of Farey fractions) he is considering the Farey sequence with $n$ large enough, so ...
Here are some of my own, they don't require iteration; just enter in an $n,$ they are extremely accurate; by some small n it should already be greater than your calculator will display and increasing to limit of $0$ error at infinity, and they meet your criteria. $\sqrt{n}\sqrt{(n+1)}\approx\frac{2n(n+1)}{2n+1}+\frac{2n+1}{4(2n+1)^{2}+1}.$ ...