= Solution
For a nonempty <set> $S\subseteq\mathbb R$, an <upper bound in a partially ordered set> is a real number $u$ such that $s\leq u$ for every $s\in S$. Its <supremum>, or least upper bound, is an upper bound $L$ with $L\leq u$ for every upper bound $u$. The <least-upper-bound property> asserts that every nonempty set of <real numbers> bounded above has a real <supremum>.
Let $(x_n)$ be a bounded, increasing <sequence>; here increasing allows $x_{n+1}=x_n$. Apply the <least-upper-bound property> to its set of terms and put $L=\sup\{x_n:n\geq1\}$. Given $\varepsilon>0$, the number $L-\varepsilon$ cannot be an upper bound, so some $x_N$ satisfies $x_N>L-\varepsilon$. The <monotone sequence> property then gives
$$
L-\varepsilon<x_N\leq x_n\leq L\qquad(n\geq N).
$$
Hence $|x_n-L|<\varepsilon$ for all $n\geq N$. This proves the <monotone bounded sequence> result, with \b[limit] $\boxed{x_n\longrightarrow L}$.
For the two <series>, define their partial sums $A_N=\sum_{n=1}^N a_n$ and $B_N=\sum_{n=1}^N b_n$. Positivity makes both partial-sum sequences increasing, and termwise comparison gives $0<B_N\leq A_N$. Since $(A_N)$ is a <convergent sequence>, it is bounded above, and therefore so is $(B_N)$. Applying the <monotone bounded sequence> result already proved shows that $(B_N)$ converges. By the definition of a <convergent series>, \b[$\sum_{n=1}^\infty b_n$ converges]. This is the <comparison test for series> for positive terms.
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