I'm too tired to work out a closed form, but... f_n = (1 - f_(n-1)) * P_n^(-1) + f_(n-1), where f_n is the fraction divisible by the first n primes and P_n is

I'm too tired to work out a closed form, but... f_n = (1 - f_(n-1)) * P_n^(-1) + f_(n-1), where f_n is the fraction divisible by the first n primes and P_n is the nth prime. So we can see the faster f_n approaches 1, the faster the first term approaches 0, and the faster f_n ~ f_(n-1). Perhaps doubly so, as if f_n approaches 1 faster, then P_n is smaller and it's reciprocal bigger as well. (We could plug in an estimate for P_n if we were converting to a closed form, to work out the asymptotic implications.)