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

All three examples below have positive integer coefficients, so they work under either convention for whether $0\in \mathbb{N}$.

For $R=1$, take $\boxed{a_n=1}$. This is the geometric series: it converges for $|z|<1$, and its terms fail to tend to zero for $|z|\ge 1$.

For $R=0$, take $\boxed{a_n=n!}$, using the factorial and $0!=1$. For every fixed $z\ne 0$, the absolute values of successive terms have ratio $(n+1)|z|$. This eventually exceeds two, so the terms grow without bound and the series cannot converge. At $z=0$ it is finite, giving radius of convergence zero.

For $R=1/\sqrt{2}$, takeusing the floor function. The coefficients satisfyThe upper bound and the comparison test for series give absolute convergence when $\sqrt{2}|z|<1$. The lower bound prevents the terms tending to zero when $\sqrt{2}|z|\ge 1$. Hence the radius of convergence is exactly $1/\sqrt{2}$.

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 $\alpha$.