Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 1 / 4D (Analysis I)

For the power series coefficient $c_n=n^2$, the Cauchy-Hadamard theorem givesThus the radius of convergence is $\boxed{R=1}$. The power series converges absolutely for $|z|<1$ and diverges for $|z|>1$. On $|z|=1$ its terms have modulus $n^2$, so they do not tend to zero and the series diverges there as well.

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.