Past exam of the mathematics course of the University of Cambridge 2001 ia Paper 1 9D iii Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2007 ia Paper 4 7E Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2008 ia Paper 1 12E ii Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2008 ia Paper 1 3F Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2010 ia Paper 1 11D Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2016 ia Paper 1 4F Solution Created 2026-09-24 Updated 2026-10-06
Every counted point has integer coordinates and lies in the square , 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 , convergence of implies convergence of . Equivalently, divergence of the smaller nonnegative series forces divergence of the larger one.
For ,For , the ratio of successive terms on the right tends to , so the ratio test and the comparison test for series show absolute convergence. The case is immediate.
If , then , so the terms do not tend to zero. The term test for divergence therefore rules out convergence, including every point on the boundary . 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.
Past exam of the mathematics course of the University of Cambridge 2017 ia Paper 1 11F b Solution Created 2026-09-24 Updated 2026-10-03
If , the terms do not tend to zero, so the series diverges by the term test for divergence. If , the terms form a nonnegative decreasing sequence, and the Cauchy condensation test gives the condensed seriesThis geometric series converges exactly when . Hence the p-series criterion is
Past exam of the mathematics course of the University of Cambridge 2019 ia Paper 1 4F ii Solution Created 2026-09-24 Updated 2026-09-29
The inequality gives, for ,For , the terms have magnitude at least and therefore do not tend to zero, so the series diverges by the term test for divergence. Hence
Suppose real coefficients satisfy for all sufficiently large , where and . Then the power series has radius of convergence one. For , the comparison test for series and the ratio test dominate it by a polynomial times a geometric sequence. For , the terms have modulus at least eventually and violate the term test for divergence. Finite changes to the coefficients do not affect the conclusion.