Method 1 — Contradiction

Solution.

First, note for \(n\in\mathbb{N}\): \[\frac{2n-1}{n+1} = \frac{2n+2-3}{n+1} = 2 - \frac{3}{n+1} < 2.\] Hence \(B\) is bounded above by \(2\). Therefore, by the completeness axiom, as \(B \neq \emptyset,\) \(\sup(B)\) exists and \(\sup(B) \leq 2.\)

Next, suppose for contradiction that \(\sup(B) < 2\). Now, for any \(x < 2,\) \[2 - \frac{3}{n+1} > x \Leftrightarrow n+1 > \frac{3}{2-x} \Leftrightarrow n > \frac{3}{2-x} - 1.\] Taking \(x = \sup(B)\) and applying Archimedes Postulate, \(\exists N \in \mathbb{N}\) such that \[\begin{align*} N &> \frac{3}{2-\sup(B)} - 1,\\ \Leftrightarrow 2 - \frac{3}{N+1} &> \sup(B), \end{align*}\] which is a contradiction as \(2 - \frac{3}{N+1} \in B.\) Hence \(\sup(B) \geq 2\), and by combining our found inequalities, \(\sup(B)=2\).