 Solution

For a nonempty set $S\subseteq \mathbb{R}$, an upper bound in a partially ordered set is a real number $u$ such that $s\le u$ for every $s\in S$. Its supremum, or least upper bound, is an upper bound $L$ with $L\le u$ 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 $L=\sup \{x_n:n\ge 1\}$. Given $\epsilon >0$, the number $L-\epsilon$ cannot be an upper bound, so some $x_N$ satisfies $x_N>L-\epsilon$. The monotone sequence property then givesHence $|x_n-L|<\epsilon$ for all $n\ge N$. This proves the monotone bounded sequence result, with limit $\boxed{x_n⟶L}$.

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 $0<B_N\le A_N$. 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, ${\sum }_{n=1}^{\infty }{b}_n$ converges. This is the comparison test for series for positive terms.