 Solution

Write $ϵ=1-q$ and $b/a=1-s$. Then $\log (1-s)=-s+O(s^2)$ and $\log q=-ϵ+O({ϵ}^2)$, givingThe largest permitted integer is still selected by the strict inequality from part (b). This says that longer master sequences need a proportionately larger selective advantage or smaller per-digit copying error.