Past exam of the mathematics course of the University of Cambridge / 2002 / ia / Paper 1 / 9C / Solution

Use the least-upper-bound property as the least upper bound axiom: every nonempty set of real numbers bounded above has a least upper bound in $\mathbb{R}$. It implies convergence of a bounded increasing sequence: if $L$ is the supremum of its terms, then for every $\epsilon >0$ some term exceeds $L-\epsilon$, and all subsequent terms lie between that term and $L$. A bounded decreasing sequence converges by applying this argument to its negative.

The alternating series test states that if $b_n\ge 0$, $b_{n+1}\le b_n$, and $b_n\to 0$, then ${\sum }_{n\ge 1}(-1{)}^{n-1}{b}_n$ converges. To prove it, let $S_N$ denote its partial sums. The even partial sums are increasing because $S_{2m+2}-S_{2m}=b_{2m+1}-b_{2m+2}\ge 0$. The odd partial sums are decreasing because $S_{2m+3}-S_{2m+1}=-b_{2m+2}+b_{2m+3}\le 0$. Moreover,Completeness of the real numbers therefore gives a limit of a sequence for each subsequence, denoted $L_e,L_o$ respectively for the even and odd subsequences. Their difference is $S_{2m+1}-S_{2m}=b_{2m+1}\to 0$, so $L_e=L_o$. Every sufficiently late partial sum belongs to one of these two subsequences, proving convergence of the full sequence of partial sums. Multiplying the terms by $-1$ leaves convergence unchanged.