Monotonic sequences (both increasing and decreasing) enjoy a very important property: they always have a limit, finite or infinite. This result, known as the monotonic sequence limit theorem, tells us precisely that an increasing sequence converges to its supremum, while a decreasing sequence converges to its infimum.
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Theorem (limit of a monotonic sequence). Let \( \{ a_n \}\) be a monotonic sequence. Then it has a limit and we have
\[ \lim_{n \to \infty} a_n = \begin{cases} \sup\limits_{n \in \mathbb{N}} a_n \in \mathbb{ R } \cup \{ +\infty \} &\text{if} \ \{ a_n \} \ \text{is increasing,} \\ \inf\limits_{n \in \mathbb{N}} a_n \in \mathbb{ R } \cup \{ -\infty \} &\text{if} \ \{ a_n \} \ \text{is decreasing.} \end{cases} \]
Proof for Increasing \( a_n \)
Proof (\( \{ a_n \} \) increasing). Let \(S = \sup\limits_{n \in \mathbb{N}} a_n\). By definition of supremum:
\[\forall n \in \mathbb{N} \quad : \quad a_n \leq S\]
\[\forall \varepsilon > 0\ \quad \exists k \in \mathbb{N} \quad : \quad S - \varepsilon < a_k\]
Since the sequence is increasing, for every \(n \geq k\) we have:
\[S - \varepsilon < a_k \leq a_n \leq S\]
Therefore:
\[ |a_n - S| < \varepsilon \quad \forall n \geq k \]
Hence, \(\lim_{n \to \infty} a_n = S = \sup\limits_{n \in \mathbb{N}} a_n\). If instead \(S = +\infty\), then \(\{a_n\}\) has no upper bounds and therefore
\[\forall M > 0 \quad \exists \nu \in \mathbb{N} \quad : \quad a_\nu > M;\]
by the increasing nature of \(\{a_n\}\) it follows that
\[a_n \geq a_\nu > M \quad \forall n \geq \nu \]
that is, \(a_n \to +\infty\) as \(n \to +\infty\).
Proof for Decreasing \( a_n \)
Proof (\( \{ a_n \} \) decreasing). Let \(L = \inf\limits_{n \in \mathbb{N}} a_n\). By definition of infimum:
\[ \forall n \in \mathbb{N} \quad : \quad L \leq a_n \]
\[ \forall \varepsilon > 0 \quad \exists k \in \mathbb{N} \quad : \quad L \leq a_k < L + \varepsilon \]
Since the sequence is decreasing, for every \(n \geq k\) we have:
\[ L \leq a_n \leq a_k < L + \varepsilon \]
Therefore:
\[ |a_n - L| < \varepsilon \quad \forall n \geq k \]
Hence, \(\lim_{n \to \infty} a_n = L = \inf\limits_{n \in \mathbb{N}} a_n\). If instead \(L = -\infty\), then \(\{a_n\}\) has no lower bounds and therefore
\[\forall M > 0 \quad \exists \nu \in \mathbb{N} \quad : \quad a_\nu < -M;\]
by the decreasing nature of \(\{a_n\}\) it follows that
\[a_n \leq a_\nu < -M \quad \forall n \geq \nu \]
that is, \(a_n \to -\infty\) as \(n \to +\infty\).
In both cases, we have proved that the limit exists and equals the supremum in the increasing case and the infimum in the decreasing case.