Monotone sequences and Cauchy sequences

 Deciding whether a given sequence $\{a_n\}$ converges or diverges is usually very difficult. Sometimes it is possible to decide if a sequence converges without knowing its limit explicitly if certain conditions are met.

Definition (Monotone increasing/decreasing sequences)

Let $\{a_n\}$ be a sequence. If

$$a_1\le a_2\le a_3\le \cdots \le a_n\le a_{n+1}\le \cdots$$

holds, that is, 

$$\mathrm{\forall }n\in \mathbb{N}(a_n\le a_{n+1}),$$

then $\{a_n\}$ is said to be monotone increasing.

If

$$a_1\ge a_2\ge a_3\ge \cdots \ge a_n\ge a_{n+1}\ge \cdots$$

hold, that is,

$$\mathrm{\forall }n\in \mathbb{N}(a_n\ge a_{n+1}),$$

then $\{a_n\}$ is said to be monotone decreasing.

If a sequence is either monotone increasing or monotone decreasing, we may simply say it is monotone.

Example. Let us define the sequence $\{a_n\}$ by

$$\begin{array}{rcl}{a}_1& =& 2,\\ a_{n+1}& =& \frac{1}{2}(a_n+\frac{2}{a_n}),n=1,2,\cdots .\end{array}$$

It can be shown (exercise!) that this sequence has the following properties:

1. $\mathrm{\forall }n\in \mathbb{N}(a_n\ge \sqrt{2})$.
2. $\{a_n\}$ is monotone decreasing.
3. $\underset{n\to \mathrm{\infty }}{lim}{a}_n=\sqrt{2}$.

Here are the first few terms.

$$\begin{array}{rcl}{a}_1& =& 2,\\ a_2& =& =\frac{1}{2}(a_1+\frac{2}{a_1})=\frac{3}{2}=1.5,\\ a_3& =& =\frac{1}{2}(a_2+\frac{2}{a_2})=\frac{17}{12}=1.41666666\cdots ,\\ a_4& =& \frac{1}{2}(a_3+\frac{2}{a_3})=\frac{577}{408}=1.41421568\cdots ,\\ a_5& =& \frac{1}{2}(a_4+\frac{2}{a_4})=\frac{665857}{470832}=1.41421356\cdots \end{array}$$

This sequence is monotone decreasing but it cannot be arbitrarily small because it is also bounded below. Therefore it must converge to some number (namely $\sqrt{2}$ in this case). That such a real number exists is guaranteed by the continuity of real numbers. □

The observation in the above example can be generalized.

Theorem (Monotone convergence theorem)

Any bounded monotone sequence converges.

Proof. Suppose $\{a_n\}$ is a bounded monotone increasing sequence. Then the set $S=\{a_n\mid n\in \mathbb{N}\}$ is bounded above so that its supremum $\alpha$ exists. For any $\epsilon >0$, $\alpha -\epsilon$ is not an upper bound of $S$ so there exists $N\in \mathbb{N}$ such that $\alpha -\epsilon <a_N$. Since $\{a_n\}$ is monotone increasing, for all $n\ge N$, $a_n\ge a_N>\alpha -\epsilon$. Since $\alpha$ is the supremum of $\{a_n\}$, we have $a_n\le \alpha <\alpha +\epsilon$. Thus we have $\alpha -\epsilon <a_n<\alpha +\epsilon$ or $|a_n-\alpha |<\epsilon$ for all $n\ge N$. Hence $\underset{n\to \mathrm{\infty }}{lim}{a}_n=\alpha$.

We can prove similarly for the case of a bounded monotone decreasing sequence. ■

Definition (Cauchy sequence)

The sequence $\{a_n\}$ is said to be a Cauchy sequence if and only if the following condition is met.

• For any $\epsilon >0$, there exists $N\in \mathbb{N}$ such that for any $k,l\ge N$, $|a_k-a_l|<\epsilon$.

Or, in a logical form,

$$\mathrm{\forall }\epsilon >0,\mathrm{\exists }N\in \mathbb{N},\mathrm{\forall }k,l\in \mathbb{N}(k,l\ge N⟹|a_k-a_l|<\epsilon ).$$

Theorem

Any convergent sequence is a Cauchy sequence.

Proof. Suppose $\underset{n\to \mathrm{\infty }}{lim}{a}_n=\alpha$. For any $\epsilon >0$, there exists $N\in \mathbb{N}$ such that for all $k,l\ge N$,

$$|a_k-\alpha |<\epsilon ,|a_l-\alpha |<\epsilon .$$

Thus,

$$|a_k-a_l|=|(a_k-\alpha )+(\alpha -a_l)|\le |a_k-\alpha |+|\alpha -a_l|<\epsilon +\epsilon =2\epsilon .$$

$2\epsilon$ is an arbitrary positive real number. Therefore, $\{a_n\}$ is a Cauchy sequence. ■

The converse is also true, but the proof is beyond the scope of this lecture.

Theorem

Any Cauchy sequence converges. 

Corollary

A sequence converges if and only if it is a Cauchy sequence.

For later convenience, we provide the following theorem without proof.

Theorem (Bolzano-Weierstrass Theorem)

Let $\{a_n\}$ be a sequence such that $a_n\in [c,d]$ for all $n\in \mathbb{N}$. Then there exists a subsequence $\{a_{n_k}\}$ of $\{a_n\}$ that converges to a value in the closed interval $[c,d]$.

Proof. Exercise. (Hint: use the squeeze theorem) ■