Thus, do there exist $n$ distinct primes whose summed reciprocals fall short of $1$ by the reciprocal of their product, for some $n\geqslant6$? I can get as far as $n=5$: $$\dfrac{1}{2}=1-\dfrac{1}{2},$$ $$\dfrac{1}{2}+\dfrac{1}{3}=1-\dfrac{1}{2\cdot3},$$ $$\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{7}=1-\dfrac{1}{2\cdot3\cdot7},$$ $$\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{1}{43}=1-\dfrac{1}{2\cdot3\cdot7\cdot43},$$ $$\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{11}+\dfrac{1}{23}+\dfrac{1}{31}=1-\dfrac{1}{2\cdot3\cdot11\cdot23\cdot31}.$$ But I can't see the way beyond that. If there is a fraction for some $n\geqslant6$, then one may naturally ask: is there a bound on such $n$?