Method 3 — Limits

Note that this doesn’t work in general, but it might be quicker when you can use it. It relies on the following theorem (which we’ll eventually cover):

Theorem 1:

A bounded, increasing sequence \((b_n)_{n \in \mathbb{N}}\) is convergent, and its limit is given by \[\lim_{n \to \infty} b_n = \sup\lbrace b_n \,\lvert\, n \in \mathbb{N} \rbrace.\]

Solution.

Define \(b_n = \frac{2n - 1}{n+1}\) for \(n \in \mathbb{N}\).

Step 1 — Show \((b_n)_n\) is bounded above:

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

Step 2 — Show \((b_n)_n\) is increasing (i.e. show \(b_{n+1} \geq b_n \; \forall n \in \mathbb{N}\)):

We have for \(n \in \mathbb{N}\), \[\begin{align*} b_{n+1} - b_{n} &= 2 - \frac{3}{n+2} - \left(2 - \frac{3}{n+1}\right),\\ &= \frac{3(n+2)-3(n+1)}{(n+1)(n+2)},\\ &= \frac{3}{(n+1)(n+2)},\\ &\geq 0. \end{align*}\] So \((b_n)\) is increasing. Hence, by the above theorem, \((b_n)\) converges, and by the Algebra of Limits, \[\sup(B) = \lim_{n \to \infty} b_n = \lim_{n \to \infty} \left(2 - \frac{\frac{3}{n}}{1 + \frac{1}{n}}\right) = 2,\] as expected!