The Monotone Convergence Theorem
If a sequence is monotone and bounded, then it converges. In other words
  1. If \((x_n)_n\) is an increasing sequence and bounded above, then \(\lim\limits_{n \rightarrow \infty} x_n\) exits and is equal to \(\sup x_n\).
  2. If \((x_n)_n\) is a decreasing sequence and bounded below, then \(\lim\limits_{n \rightarrow \infty} x_n\) exits and is equal to \(\inf x_n\).

Notes

Why do we need this? So far every sequence we saw has a general term. For example the sequence \(x_n = \frac{1}{n}\) or \(x_n = (-1^n)\). But in general, a lot of sequences that we will encounter don’t have a general term. We will see something like \(x_n = x_{n-1} + x_{n-2}\) where the current term depends on previous terms. How do we find the limit of such sequences? We can’t apply the definition of limit directly since it will be useless. Then, what do we do? The monotone convergence theorem is one tool that we can use to study such sequences!

Proof

Let \((x_n)_n\) be a sequence that is increasing and bounded above. We know that \((x_n)\) is bounded so if we consider the set

$$ \begin{align*} S = \{x_n : n \in \mathbb{N}\} \end{align*} $$

Then, we claim that \(\lim\limits_{n \rightarrow \infty} x_n = \sup S\). To show this, let \(\epsilon > 0\) be given. We want to show that there exists some \(N \in \mathbb{N}\) such that when \(n \geq N\), we must have

$$ \begin{align*} \lvert x_n - \sup S \rvert < \epsilon. \end{align*} $$

By definition, \(s - \epsilon\) is not a supremum so we know that there exists some term \(x_N \in S\) such that

$$ \begin{align*} \sup S - \epsilon \leq x_N \end{align*} $$

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

$$ \begin{align*} x_N \leq x_n. \end{align*} $$

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

$$ \begin{align*} \sup S - \epsilon &\leq x_N \leq x_n \leq s \leq \sup S + \epsilon \end{align*} $$

From this we see that,

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

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

References