3.5 The Cauchy Criterion

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Summary

Definition

• Cauchy sequence
• Contractive sequence

Theorem

• $X$ sequence of real numbers

1. $X$ convergent $\leftrightarrow$ Cauchy $\to$ bounded
2. $X$ contractive $\to$
   • Cauchy
   • 3.5.10

3.5.1 Definition: Cauchy sequence

A sequence $X=(x_n)$ of real numbers is said to be a Cauchy sequence

if for every $\epsilon >0$ there exists $H(\epsilon )\in \mathbb{N}$ such that for all natural numbers $n,m\ge H(\epsilon )$, the terms $x_n,x_m$ satisfy $|x_n-x_m|<\epsilon$.

• Cauchy sequence:

  |$x_n-x_m$| converges 

• Not Cauchy sequence

  

3.5.3 Lemma

If $X=(x_n)$ is a convergent sequence of real numbers, then $X$ is a Cauchy sequence.

NOTE

$X$ convergent and Cauchy is a biimplication as stated on 3.5.5

3.5.4 Lemma

A Cauchy sequence of real numbers is bounded.

3.5.5 Thoerem: Cauchy Converge Criterion

A sequence of real numbers is convergent if and only if it is a Cauchy sequence.

3.5.7 Definition: Contractive sequence

We say that a sequence $X=(x_n)$ of real numbers is contractive if there exists a constant $C,0<c<1$, such that $|x_{n+2}-x_{n+1}|\le C|x_{n+1}-x_n|,\forall n\in \mathbb{N}$ The number $C$ is called the constant of the contractive sequence.


3.5.8 Theorem

Every contractive sequence is a Cauchy sequence, and therefore is convergent.

3.5.10 Corollary

If $X:=(x_n)$ is a contractive sequence with constant $C,0<C<1$, and if $x^{\ast }:=\lim X$, then

1. $|x^{\ast }-x_n|\le \frac{C^{n-1}}{1-C}|x_2-x_1|$
2. $|x^{\ast }-x_n|\le \frac{C}{1-C}|x_n-x_{n-1}|$

Proofs

3.5.3

Direct proof.

1. Premise

Let $x:=\lim X$ and $\epsilon >0$ be given.

2. From 1. $X$ has limit → $X$ is convergent. Use Definition of convergence

There exists $K(\epsilon /2)\in \mathbb{N}$ such that if $n\ge K(\epsilon /2)$ then $|x_n-x|<\epsilon /2$.

3. From 2, Triangle inequality. Form 3.5.1 Definition: Cauchy sequence

Thus, if $H(\epsilon ):=K(\epsilon /2)$ and if $n,m\ge H(\epsilon )$, then we have

$$\begin{array}{rl}|x_n-x_m|& =|(x_n-x)+(x-x_m)|\\ & \le |x_n-x|+|x_m-x|<\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon \end{array}$$

4. Conclusion

Since $\epsilon >0$ is arbitrary, it follows that $(x_n)$ is a Cauchy sequence.

3.5.4

Direct proof.

1. Premise

Let $X:=(x_n)$ be a Cauchy sequence and let $\epsilon :=1$.

2. Cauchy definition. $m=H$

If $H:=H(1)$ and $n\ge H$, then $|x_n-x_H|<1$.

3. $|a-b|\le |a|-|b|$

Hence, by Triangle Inequality, we have $|x_n|\le |x_H|+1\forall n\ge H$.

4. From 3. Forming Definition: Boundedness of a sequence

If we set

$$M:=\sup \{|x_1|,|x_2|,\dots ,|x_{H-1}|,|x_H|+1\}$$

then it follows that $|x_n|\le M\forall n\in \mathbb{N}$

Example

1. Prove $(x_n):=\left(\frac{n+1}{n}\right)$ is a Cauchy sequence

Misal $(x_n)$ barisan bilangan real.

Definisikan $(x_n):=\left(\frac{n+1}{n}\right)$.

Ambil sembarang $\epsilon >0$. Berdasarkan Sifat Archimedes, $\exists H(\epsilon )\in \mathbb{N}$ sedemikian sehingga $H(\epsilon )>\frac{2}{\epsilon }$.

Akibatnya, $\forall n,m\in \mathbb{N}$ berlaku:

$$\begin{array}{rl}|x_n-x_m|& =\left|\left(1+\frac{1}{n}\right)-\left(1+\frac{1}{m}\right)\right|\\ & =\left|\frac{1}{n}-\frac{1}{m}\right|\\ & \le \left|\frac{1}{n}\right|+\left|\frac{1}{m}\right|\\ & <\frac{1}{H(\epsilon )}+\frac{1}{H(\epsilon )}\\ & =\frac{2}{H(\epsilon )}\end{array}$$

$\therefore$ $\forall \epsilon >0\exists H(\epsilon )\in \mathbb{N}\ni \forall n,m>H(\epsilon ),|x_n-x_m|<\epsilon$

$\therefore (x_n):=\left(\frac{n+1}{n}\right)$ adalah barisan Cauchy.