2.7 The monotone convergence theorem

So far, in order to show a sequence $(a_{n})_{n\in\mathbb{N}}$ converges, we have always started with a candidate limit point $a\in\mathbb{R}$ and then worked with the definition to check that, indeed, $a_{n}\to a$ as $n\to\infty$ . However, there are many instances where it is not at all obvious in advance what the limit should be. For this reason, it is useful to develop tools which allow us to determine whether a sequence converges without having to know the value of the limit.

Theorem 2.68 (Monotone convergence theorem).

Let $(a_{n})_{n\in\mathbb{N}}$ be a sequence.

1

If $(a_{n})_{n\in\mathbb{N}}$ is non-decreasing and bounded above, then $(a_{n})_{n\in\mathbb{N}}$ converges.
2

If $(a_{n})_{n\in\mathbb{N}}$ is non-increasing and bounded below, then $(a_{n})_{n\in\mathbb{N}}$ converges.

The monotone convergence theorem is a powerful result with many important consequences. We highlight two important features:

1.

The theorem allows us to deduce the existence of a limit of a sequence without having to know the precise value of the limit beforehand. Consequently, it is very useful in existence proofs and we shall see many applications, especially when we study series in the next section.
2.

We shall see below that the monotone convergence theorem is closely related to the completeness axiom. In particular, it can be thought of as bridge between the completeness axiom and the theory of sequences.

Proof (of Theorem 2.68).

We shall only prove part 1, since part 2 follows from a very similar argument (and can also be deduced from part 1: see Exercise 2.70 below).

Consider the set of terms of the sequence, $A:=\{a_{n}:n\in\mathbb{N}\}$ . Clearly $A$ is nonempty and, since the sequence $(a_{n})_{n\in\mathbb{N}}$ is bounded above, $A$ is bounded above. Thus, by the completeness axiom, $s:=\sup A$ exists. We claim $a_{n}\to s$ .

To prove the claim, let $\varepsilon>0$ be given. By the approximation property from Lemma 1.31, there exists some $N\in\mathbb{N}$ such that $s-\varepsilon\leq a_{N}\leq s$ . Since $(a_{n})_{n\in\mathbb{N}}$ is monotone non-decreasing, it follows that $s-\varepsilon\leq a_{n}\leq s$ for all $n\geq N$ . Thus, by the definition of a limit, $a_{n}\to s$ , as claimed. ∎

Remark 2.69.

Note that the proof of the theorem gives us more information: if $(a_{n})_{n\in\mathbb{N}}$ is non-decreasing and bounded above, then $(a_{n})_{n\in\mathbb{N}}$ converges and $\lim_{n\to\infty}a_{n}=\sup\{a_{n}:n\in\mathbb{N}\}$ . In particular, $a_{n}\leq\lim_{n\to\infty}a_{n}$ for all $n\in\mathbb{N}$ .

Exercise 2.70.

Use part 1 of Theorem 2.68 to prove part 2 of Theorem 2.68. Hint: Exercise 2.14 and Exercise 2.21 are useful here.

Clearly the monotone convergence theorem is closely related to the completeness axiom: the axiom plays a starring role in the proof! Moreover, we can relate the monotone convergence theorem back to our recurring theme that the real line does not contain any holes. We illustrate this by using the monotone convergence theorem to again show the existence of various real numbers which are missing from $\mathbb{Q}$ .

Exercise 2.71.

We shall give another proof showing $2$ has a square root in $\mathbb{R}$ (in particular, for the purpose of the exercise, to avoid circularity you should not assume $2$ has a square root).

(i)

Let $a_{1}:=2$ and define a sequence of real numbers $(a_{n})_{n\in\mathbb{N}}$ recursively by setting

$a_{n+1}:=a_{n}-\frac{1}{2}\Big{(}a_{n}-\frac{2}{a_{n}}\Big{)}=\frac{a_{n}}{2}% \Big{(}1+\frac{2}{a_{n}^{2}}\Big{)}\qquad\text{for all $n\in\mathbb{N}$.}$

Using induction, show that $a_{n}>0$ for all $n\in\mathbb{N}$ . In particular, conclude that the sequence is well-defined (we never divide by $0$ ).
(ii)

Show that $a_{n+1}^{2}=2+\frac{1}{4}\big{(}a_{n}-\frac{2}{a_{n}}\big{)}^{2}$ for all $n\in\mathbb{N}$ and conclude that $a_{n}^{2}\geq 2$ for all $n\in\mathbb{N}$ .
(iii)

Using the results of parts (i) and (ii), show that $(a_{n})_{n\in\mathbb{N}}$ is non-increasing and bounded below.
(iv)

Conclude that $(a_{n})_{n\in\mathbb{N}}$ converges and show that the limit $a\in\mathbb{R}$ satisfies $a^{2}=2$ .

Thus, we have shown that $2$ has a square root.

Exercise 2.71 demonstrates one of the great strengths of the monotone convergence theorem, highlighted in item 1 above: the theorem allows us to deduce that a limit exists without having to know its value. This feature makes it a powerful tool for existence proofs, as above (where we are showing the existence of square roots of positive real numbers).

Exercise 2.72.

Consider the sequence $(a_{n})_{n\in\mathbb{N}}=(0,0.3,0.33,0.333,\dots)$ . Show that $a_{n}\to a$ as $n\to\infty$ for some $a\in\mathbb{R}$ and use Exercise 2.33 to deduce that $3a=1$ . This is the rigorous interpretation of the decimal expansion $0.333\ldots=1/3$ .⁶⁶ 6 We shall study decimal expansions more systematically in the next section.

Exercise 2.73.

Let $r>0$ and $(a_{n})_{n\in\mathbb{N}}$ be given by $a_{n}:=r^{1/n}$ for all $n\in\mathbb{N}$ .

(i)

Use the monotone convergence theorem to show that $(r^{1/n})_{n\in\mathbb{N}}$ converges to some limit $L>0$ .
(ii)

Apply the subsequence test to the squared sequence $(a_{n}^{2})_{n\in\mathbb{N}}$ to show $L=L^{2}$ . Conclude that $r^{1/n}\to 1$ as $n\to\infty$ .