Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics II / 3 / c / ii / Solution

Split the power series into even and odd indices:whenever the two series converge absolutely. If $|z|<1$, the even part is a convergent geometric series, while the odd part is bounded termwise in absolute value by ${\sum }_j|z{|}^{2j+1}$, also a convergent geometric series. Hence there is absolute convergence throughout $|z|<1$.

If $|z|>1$, the odd terms have absolute values $|z{|}^{2j+1}/(2j+1{)}^2$. Their successive ratio tends to $|z{|}^2>1$, so these terms grow eventually and do not tend to zero. The term test for divergence therefore forces divergence of the original power series. Consequently its radius of convergence is