Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics II / 2 / b / Solution

Put $u_n=11^n\log n/(2^{2n+1}n!)$ for $n\ge 2$. The factorial givesThe logarithm identityfollows from continuity of the logarithm at one and $\log n\to +\infty$. The product rule for limits of sequences therefore makes the displayed ratio tend to zero. In particular, some $N$ satisfies $0<u_{n+1}\le u_n/2$ for all $n\ge N$. By mathematical induction, $0<u_{N+j}\le u_N{2}^{-j}$. A geometric sequence with ratio in $(0,1)$ tends to zero, so the squeeze theorem proves