Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics II / 3 / c / ii / Solution

Use the ceiling function to choose positive integer coefficientsBecause $0<\alpha <1$, we have ${\alpha }^{-n}\ge 1$ andFor $|z|<\alpha$, the comparison test for series with the geometric series givesFor $|z|\ge \alpha$, we instead have $|a_n{z}^n|\ge (|z|/\alpha {)}^n\ge 1$, so the terms do not tend to zero. Thus the power series has radius of convergence exactly $\alpha$.