Monotone Convergence Theorem

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

Proof
Let $(a_{n})$ be monotone and bounded. To show $(a_{n})$ converges, we need a candidate limit. Let's assume the sequence is increasing (the decreasing case is handled similarly), and consider the set of points ${a_{n} | n \in N}$ . By assumption, this set is bounded, so then:

s = sup ({a_{n} | n \in N})

exists. It seems reasonable to claim that $(a_{n}) \to s$ so let's show it via the definition of convergence of a sequence.

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Let $ε > 0$ . Since $s$ is a least upper bound for the set in question, then $s - ε$ is not an upper bound. Hence, there is a point in the sequence $a_{N}$ such that $s - ε < a_{N}$ . Now since $(a_{n})$ is increasing then any $n \geq N$ has it that $a_{N} \leq a_{n}$ . Hence:

s - ε < a_{n} \leq \underset{s is an upper bound}{\underset{⏟}{a_{n} \leq s}} < s + ε

Consequently then $| a_{n} - s | < ε$ by using the ends and the middle of the inequality.
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An alternate proof:

The idea is that if we have the set $A$ be the $a_{i}$ 's, then it must have a supremum, which must be the limit's value itself.

Proof
Let $(a_{n})$ be bounded and assume it is increasing (from the monotonic part). Let $A = {a_{n} | n \in N}$ where $A \neq \emptyset$ and $A$ is bounded. Let $a = sup (A)$ . We'll show that $a = lim a_{n}$ .

Let $ε > 0$ . Since $a$ is a suprema, then by the Suprema Lemma then $\exists a^{'} \in A$ where:

a - ε < a^{'}

But $a^{'} \in A$ , so then $\exists N \in N$ where $a^{'} = a_{N}$ . Then:

a - ε < a_{N}

Now assume $n \geq N$ . Since $(a_{n})$ is increasing. Then:

a_{n} \geq a_{N} > a - ε

By flipping the signs:

a_{n} \leq a < a + ε

That implies:

- ε < a_{n} - a < ε \Rightarrow | a_{n} - a | < ε

The decreasing version is proved in a very similar way.
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Example

Show that $(a_{n}) = x^{n}$ is convergent for all $x \in (0, 1)$ . Show it converges to $0$ .

Proof
Notice that $x^{n}$ is decreasing since $x \cdot x^{n} \leq x^{n}$ . Similarly, $x^{n}$ is bounded since it's in $(0, 1)$ . As such, the MST suggests that $\exists L$ where $lim x^{n} = L$ .

But notice also that the ALT suggests that $b_{n} \to x L$ where $b_{n} = x a_{n}$ . Notice that then $b_{n} \to L$ so then $x L = L \Rightarrow L = 0$ .
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