**Proof Plan.** We treat the increasing case; the decreasing case follows by applying the result to $\{-a_n\}$. The key idea is that the supremum of the [sequence](/page/Sequence) exists (by completeness of $\mathbb{R}$) and is the [limit](/page/Limit). **Step 1 (Identify the candidate limit).** Since $\{a_n\}$ is bounded above, the [set](/page/Set) $\{a_n : n \in \mathbb{N}\}$ has a supremum in $\mathbb{R}$ by the least upper bound property. Set $L := \sup_{n \in \mathbb{N}} a_n$. **Step 2 (Verify convergence).** Let $\varepsilon > 0$. By the characterisation of the supremum, $L - \varepsilon$ is not an upper bound of $\{a_n\}$, so there exists $N \in \mathbb{N}$ with $a_N > L - \varepsilon$. Since $\{a_n\}$ is increasing, for all $n \ge N$: \begin{align*} L - \varepsilon < a_N \le a_n \le L. \end{align*} Therefore $|a_n - L| = L - a_n < \varepsilon$ for all $n \ge N$, so $a_n \to L$. $\blacksquare$