Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics II / 2 / b / iii / Solution

True. Since $a_n^n\to 2$, there is $N_0$ such that $1<a_n^n<3$ for $n\ge N_0$. Positivity then gives $a_n>1$ on this tail.

Fix $\epsilon >0$. If $a_n\ge 1+\epsilon$, Bernoulli's inequality givesFor $n>2/\epsilon$, this contradicts $a_n^n<3$. Therefore eventually $1<a_n<1+\epsilon$, proving