Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics II / 3 / b

The Cauchy-Hadamard theorem states $R^{-1}=\mathrm{limsup}|a_n{|}^{1/n}$. At the prime numbers $n=p$, this root equals $p^{2/p}\to 1$. There are infinitely many prime numbers, so these indices form an unbounded subsequence. For every other positive index the root is zero. The limit superior is therefore one, and $\boxed{R=1}$.

By Stirling's approximation,Here $(2\pi n{)}^{1/(2n)}\to 1$ and $n^{1/n^2}\to 1$, as is seen by taking the natural logarithm. The Cauchy-Hadamard theorem gives $\boxed{R=e^{-1}}$.

At the nonzero indices $n=m!$ with $m\ge 2$,since $m/(m!)=1/(m-1)!\to 0$. At all other sufficiently large indices the coefficient is zero. The initial coefficients do not affect the limit superior, which equals one. The Cauchy-Hadamard theorem therefore gives $\boxed{R=1}$.