Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 1 / 4D / ii / Solution

The original PDF has coefficient $c_n=n^{n^{1/3}}$. Its $n$th root isThe Cauchy-Hadamard theorem therefore gives $\boxed{R=1}$. Again the power series converges absolutely inside the unit circle, while on that circle its terms do not tend to zero. The converted TeX loses the outer exponent and writes $n^{1/3}$; the calculation here uses the printed coefficient.