Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 4 / 12I / c / Solution

Here $q_{n+1}\ge a_{n+1}{q}_n=q_n^{n+1}$, so $|\alpha -p_n/q_n|<q_n^{-(n+2)}$. If $\alpha$ were algebraic of fixed degree $d$, Liouville's theorem would contradict this for $n>d$. Thus $\alpha$ is transcendental.

Solved by gpt-5.6-sol high.