Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / Paper 1 / 19 / d / Solution

A rational lower bound for the natural logarithm proves positivity for every $n$, including $n=1$. DefineThen $h(0)=0$, while direct differentiation givesThus $h(t)>0$ for $t>0$. Setting $t=1/n$ in the expression from part (b) givesTherefore $\boxed{u_n>0\text{ for every }n\ge 1}$, and $(a_n)$ is a strictly increasing sequence.