Khanh at 04:26, 22 November 2021

2021-11-22T04:26:01Z

Khanh: Created page with "Recall that a sequence of real numbers $(a_n)$ is said to be a Cauchy Sequence if $\forall \epsilon > 0$ there exists an $N \in \ma..."$

2021-11-22T04:23:44Z

Created page with "Recall that a sequence of real numbers <math>(a_n)</math> is said to be a <strong>Cauchy Sequence</strong> if <math>\forall \epsilon > 0</math> there exists an <math>N \in \ma..."

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Recall that a sequence of real numbers <math>(a_n)</math> is said to be a <strong>Cauchy Sequence</strong> if <math>\forall \epsilon > 0</math> there exists an <math>N \in \mathbb{N}</math> such that if <math>m, n \geq N</math> then <math>\mid a_n - a_m \mid < \epsilon</math>. We already noted that every convergent sequence of real numbers is Cauchy, and that every Cauchy sequence of real numbers is bounded. We will now look at another important theorem known as the '''Cauchy Convergence Criterion'''.

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
:'''Theorem (Cauchy Convergence Criterion):''' If <math>(a_n)</math> is a sequence of real numbers, then <math>(a_n)</math> is convergent if and only if <math>(a_n)</math> is a Cauchy sequence.
</blockquote>

Note that the Cauchy Convergence Criterion will allow us to determine whether a sequence of real numbers is convergent whether or not we have a suspected limit in mind for a sequence.

*'''Proof:''' <math>\Rightarrow</math> Suppose that <math>(a_n)</math> is a convergent sequence of real numbers.

* <math>\Leftarrow</math> Suppose that <math>(a_n)</math> is a Cauchy sequence. We want to show that <math>(a_n)</math> is thus convergent to some real number in <math>\mathbb{R}</math>. Now since <math>(a_n)</math> is a Cauchy sequence it follows that <math>(a_n)</math> is bounded. Since <math>(a_n)</math> is bounded, it follows from <a href="/the-bolzano-weierstrass-theorem">The Bolzano-Weierstrass Theorem</a> that there exists a subsequence of <math>(a_n)</math>, call it <math>(a_{n_k})</math> that converges to some real number <math>A^*</math>.

* Since <math>(a_n)</math> is a Cauchy sequence, then <math>\forall \epsilon > 0</math> there exists an <math>N \in \mathbb{N}</math> such that if <math>m, n \geq N</math> then <math>\mid a_n - a_m \mid < \epsilon</math>. Choose <math>\epsilon_1 = \frac{\epsilon}{2} > 0</math>, and so there exists an <math>N_1 \in \mathbb{N}</math> such that if <math>m, n \geq N_1</math> then <math>\mid a_n - a_m \mid < \epsilon_1 = \frac{\epsilon}{2}</math>.

* Now look at the subsequence <math>(a_{n_k})</math> which converges to <math>A^*</math>. There thus exists a natural number <math>K \geq N_1</math> where <math>K</math> belongs to the set of indices <math>\{ n_1, n_2, ... \}</math> such that <math>\mid a_{K} - A^* \mid < \frac{\epsilon}{2}</math>.

*Since <math>K \geq N_1</math> then if we substitute <math>m = K</math> we have that for <math>n \geq N_1</math>:

<div style="text-align: center;"><math>\begin{align} \mid a_n - a_K \mid < \frac{\epsilon}{2} \end{align}</math></div>

*And so for <math>n \geq N_1</math> we have that:

<div style="text-align: center;"><math>\begin{align} \quad \mid a_n - A^* \mid = \mid a_n - a_K + a_K - A^* \mid \leq \mid a_n - a_K \mid + \mid a_K - A^* \mid < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}</math></div>

*Therefore <math>\lim_{n \to \infty} a_n = A^*</math>, and so <math>(a_n)</math> is convergent to the real number <math>A^*</math>. <math>\blacksquare</math>

We will summarize the lemma from the Cauchy Sequences and the Cauchy Convergent Criterion as follows:

*Any sequence of real numbers <math>(a_n)</math> that is convergent is also a Cauchy sequence.

*Any Cauchy sequence of real numbers <math>(a_n)</math> is also a convergent sequence.

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	*Any Cauchy sequence of real numbers <math>(a_n)</math> is also a convergent sequence.		*Any Cauchy sequence of real numbers <math>(a_n)</math> is also a convergent sequence.
		+
		+	== Licensing ==
		+	Content obtained and/or adapted from:
		+	* [http://mathonline.wikidot.com/the-cauchy-convergence-criterion The Cauchy Convergence Criterion, mathonline.wikidot.com] under a CC BY-SA license

The Cauchy Criterion for Convergence - Revision history

Khanh at 04:26, 22 November 2021

Khanh: Created page with "Recall that a sequence of real numbers (a_n) is said to be a Cauchy Sequence if \forall \epsilon > 0 there exists an N \in \ma..."

Khanh: Created page with "Recall that a sequence of real numbers $(a_n)$ is said to be a Cauchy Sequence if $\forall \epsilon > 0$ there exists an $N \in \ma..."$