If a sequence is monotone and bounded, then it converges.


For the absolute value function definition and other properties see here.
For the definitions of sequences and what it means to for a sequence to converge, see this.
For the “show the limit” template and an example, see this.

Proof:

Let \((a_n)\) be a sequence that is monotone and bounded. Suppose without the loss of generality that \((a_n)\) is increasing. We know that \((a_n)\) is bounded so if we consider the set \(\{a_n : n \in \mathbf{N}\}\), then we can let

$$ \begin{align*} s = \sup{\{a_n : n \in \mathbf{N}\}}. \end{align*} $$

The claim is \((a_n)\) converges to \(s\) or \(\lim(a_n) = s\). To prove that, we need to show that for any \(\epsilon > 0\), there exists some \(n \in \mathbf{N}\) such that,

$$ \begin{align*} \lvert a_n - s \rvert < \epsilon \quad \text{whenever $n \geq N$.} \end{align*} $$

Let \(\epsilon > 0\). First, by lemma 1.3.8, \(s - \epsilon\) is not an upper bound so we know that there exists some \(a_N\) such that

$$ \begin{align*} s - \epsilon \leq a_N. \end{align*} $$

Second, since \((a_n)\) is increasing, then if we let \(n \geq N\), we will have

$$ \begin{align*} a_N \leq a_n. \end{align*} $$

Combining both inequalities and adding the fact \(s + \epsilon\) is an upper bound on the set, then

$$ \begin{align*} s - \epsilon &\leq a_N \leq a_n \leq s \leq s + \epsilon \end{align*} $$

From this we see that,

$$ \begin{align*} s - \epsilon &\leq a_n \leq s + \epsilon \quad \text{whenever $n \geq N$}. \\ \end{align*} $$

Therefore, \(\lvert a_n - s \rvert < \epsilon\) and so \(\lim(a_n) = s\) as we wanted to show. \(\blacksquare\)

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