 Solution

A family exhibits pre-cutoff for Markov chains if there are constants $0<A<B<\infty$ and times $t_n$ such that its worst-case total-variation distance tends to one at $A{t}_n$ and to zero at $B{t}_n$. For irreducible reversible chains, a standard necessary condition for pre-cutoff is the product conditionThe hypothesis that $t_{\mathrm{mix}}^{(n)}/t_{\mathrm{rel}}^{(n)}$ is bounded contradicts this condition, while $t_{\mathrm{mix}}^{(n)}\to \infty$ excludes a bounded-time degeneracy. Hence the family cannot exhibit pre-cutoff.