Past exam of the mathematics course of the University of Cambridge 2002 ia Paper 1 9C i Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2004 ia Paper 1 12F iii Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2009 ia Paper 1 4E Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2009 ia Paper 1 9F a Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2009 ia Paper 1 9F b Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2010 ia Paper 1 9E c Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2010 ia Paper 1 9E d Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2013 ia Paper 1 4F a Solution Created 2026-09-24 Updated 2026-10-07
Let . Pairing consecutive terms shows andThe even partial sums increase and the odd partial sums decrease. They satisfy , whileBoth are monotone bounded sequences and hence have limits, and the displayed difference makes their limits equal. Every partial sum belongs to one subsequence, so the alternating series converges. This is a proof of the alternating series test; the monotonicity and zero-limit hypotheses play separate roles.
Past exam of the mathematics course of the University of Cambridge 2021 ia Paper 1 3F Solution Created 2026-09-24 Updated 2026-09-25
To prove it, let be the partial sums. The even sums satisfyso is increasing. The odd sums satisfyso is decreasing. Also , so both are bounded and converge. Their difference is , hence their limits agree and the whole sequence converges.
Taking proves convergence of the alternating harmonic series. Its even partial sums lie below its limit , while its odd partial sums lie above it. Sinceandwe obtain
Past exam of the mathematics course of the University of Cambridge 2022 ia Paper 1 3D Solution Created 2026-09-23 Updated 2026-09-25
The alternating series test says that if decreases to , thenconverges. Taking proves convergence of the given series.
It is not absolute convergence, because the series of absolute values is the p-serieswhich diverges since . Thus the original series has conditional convergence.
For an explicit divergent rearrangement of a series, take unused positive terms, which are the even-indexed terms, until the partial sum exceeds , then take the first unused negative term. Next take positive terms until the sum exceeds , then the next unused negative term, and continue. Both the positive and negative subseries have infinite total magnitude, so this procedure uses every term. The negative term inserted at stage tends to zero, while the preceding partial sum exceeds ; consequently these rearranged partial sums tend to . This is a divergent series with exactly the prescribed terms.