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

The radius of convergence of a power series ${\sum }_{n=0}^{\infty }{a}_n{z}^n$ is the number $R\in [0,\infty ]$ such that the series has absolute convergence for $|z|<R$ and diverges for $|z|>R$. Equivalently,Use $r^0=1$, including at $r=0$. The supremum may be $+\infty$. There is no general conclusion at $|z|=R$: boundary convergence depends on the coefficients. If $R=0$, only the centre is guaranteed; if $R=\infty$, every finite $z$ lies inside the disk of convergence.

The comparison test for series states that, for $0\le u_n\le v_n$ eventually, convergence of $\sum v_n$ implies convergence of $\sum u_n$. Equivalently, divergence of $\sum u_n$ implies divergence of $\sum v_n$.

For $|z|<1$, the coefficient bound givesby the geometric series formula and the comparison test for series. The single term $a_0$ does not affect convergence. For $|z|\ge 1$,so the terms do not tend to zero. The fact that terms of a convergent series tend to zero excludes convergence there. Thus

If $z_0=0$, there is no $z$ with $|z|<|z_0|$, so the claim is vacuous. Suppose $z_0\ne 0$. Convergence of $\sum a_n{z}_0^n$ implies $a_n{z}_0^n\to 0$, because terms of a convergent series tend to zero. In particular this is a bounded sequence, so $|a_n{z}_0^n|\le C$ for some finite $C$ and every $n$.

For $|z|<|z_0|$, put $q=|z|/|z_0|<1$. ThenThe geometric series and comparison test for series proveThus convergence at one point gives absolute convergence at every point strictly closer to the centre.