= 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 $\varepsilon>0$ some term exceeds $L-\varepsilon$, 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 $b_n\ge0$, $b_{n+1}\le b_n$, and $b_n\to0$, then $\sum_{n\ge1}(-1)^{n-1}b_n$ converges. To prove it, let $S_N$ denote its <partial sums>. The even <partial sums> are increasing because $S_{2m+2}-S_{2m}=b_{2m+1}-b_{2m+2}\ge0$. The odd <partial sums> are decreasing because $S_{2m+3}-S_{2m+1}=-b_{2m+2}+b_{2m+3}\le0$. Moreover,
$$
0\le S_{2m}\le S_{2m+1}\le b_1.
$$
<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 $S_{2m+1}-S_{2m}=b_{2m+1}\to0$, 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.
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