Past exam of the mathematics course of the University of Cambridge / 2016 / ia / Paper 1 / 4F / Solution

Every counted point has integer coordinates and lies in the square $[-n,n{]}^2$, while the origin is always counted. HenceNo precise circle-area estimate is needed; this elementary bound controls the growth of the power series coefficients.

The comparison test for series states that if $0\le u_n\le v_n$, convergence of $\sum v_n$ implies convergence of $\sum u_n$. Equivalently, divergence of the smaller nonnegative series forces divergence of the larger one.

For $|z|=r<1$,For $0<r<1$, the ratio of successive terms on the right tends to $r<1$, so the ratio test and the comparison test for series show absolute convergence. The case $r=0$ is immediate.

If $|z|\ge 1$, then $|a_n{z}^n|\ge 1$, so the terms do not tend to zero. The term test for divergence therefore rules out convergence, including every point on the boundary $|z|=1$. Thus the radius of convergence isThe series converges exactly in the open unit disc. This is an example of polynomial coefficient growth with unit power-series radius: the lattice-count bound is polynomial rather than exponential.