Divergence Criteria - Revision history

Khanh at 20:12, 7 November 2021

2021-11-07T20:12:29Z

Lila at 19:11, 20 October 2021

2021-10-20T19:11:44Z

Lila at 19:10, 20 October 2021

2021-10-20T19:10:42Z

Lila at 19:09, 20 October 2021

2021-10-20T19:09:46Z

Lila at 19:08, 20 October 2021

2021-10-20T19:08:53Z

Lila: Created page with "

The Divergence Criteria for Sequences

Thus far we have looked at criteria for sequences to be convergent. We will now begin to look at some criteria whic..."

2021-10-20T19:07:14Z

Created page with "<h1 id="toc0">The Divergence Criteria for Sequences</h1> <p>Thus far we have looked at criteria for sequences to be convergent. We will now begin to look at some criteria whic..."

New page

<h1 id="toc0">The Divergence Criteria for Sequences</h1>
<p>Thus far we have looked at criteria for sequences to be convergent. We will now begin to look at some criteria which will tell us if a sequence is divergent.</p>
<table class="wiki-content-table">
<tr>
<td><strong>Theorem 1 (Divergence Criteria for Sequences):</strong> A sequence <math>(a_n)</math> of real numbers is divergent if either one of the following hold:<br />
<strong>1.</strong> <math>(a_n)</math> has two subsequences <math>(a_{n_k})</math> and <math>(a_{n_p})</math> that converge to two different limits.<br />
<strong>2.</strong> <math>(a_n)</math> has a subsequence that is divergent.<br />
<strong>3.</strong> <math>(a_n)</math> is unbounded.</td>
</tr>
</table>
<p>Notice that if either <strong>(1)</strong> or <strong>(2)</strong> hold then this immediately contradicts the fact that if a sequence <math>(a_n)</math> is convergent then all of its subsequences <math>(a_{n_k})</math> converge to the same limit.</p>
<p>Recall by <a href="/the-boundedness-of-convergent-sequences-theorem">The Boundedness of Convergent Sequences Theorem</a> that if a sequence is convergent that it is bounded. The contrapositive of this statement is that is a sequence is not bounded then it is divergent, and so then <strong>(3)</strong> is justified as well.</p>
<p>We will now look at some examples of apply the Divergence Criteria for Sequences.</p>
<h2 id="toc1">Example 1</h2>
<p><strong>Show that the sequence <math>((-1)^n)</math> is divergent.</strong></p>
<p>To show this sequence is divergent, consider the subsequence of even terms which is <math>(1, 1, 1, ... )</math> which converges to the real number <math>1</math>. Now consider the subsequence of odd terms which is <math>(-1, -1, -1, ... )</math> which converges to the real number <math>-1</math>.</p>
<p>Since these two subsequences converge to different limits, we conclude that <math>((-1)^n)</math> is divergent.</p>
<h2 id="toc2">Example 2</h2>
<p><strong>Show that the sequence <math>(a_n)</math> defined by <math>a_n = \left\{\begin{matrix} 1/n & \mathrm{if \: n \: is \: even} \\ n & \mathrm{if \: n \: is \: odd}\end{matrix}\right.</math>.</strong></p>
<p>Let's first look at a few terms of this sequence. We have that <math>(a_n) = \left(1, \frac{1}{2}, 3, \frac{1}{4}, 5, \frac{1}{6}, ... \right)</math>. We can see this sequence is not bounded above and hence not bounded, which we will prove.</p>
<p>Suppose that there exists an <math>M \in \mathbb{N}</math> such that <math>\mid a_n \mid M</math> for all <math>n \in \mathbb{N}</math>. By the Archimedean property since <math>M \in \mathbb{R}</math> there exists a natural number <math>n_M \in \mathbb{N}</math> such that <math>M \leq n_M</math>. We also note that <math>M \leq n_M n_M + 1</math>. So either <math>n_M</math> or <math>n_M + 1</math> is a term in the sequence <math>(a_n)</math> which contradicts the sequence <math>(a_n)</math> from being bounded.</p>
<h2 id="toc3">Example 3</h2>
<p><strong>Show that the sequence <math>(n)</math> is divergent.</strong></p>
<p>Once again, this sequence is unbounded. Suppose that instead <math>(n)</math> is bounded, that is <math>\mid n \mid = n M</math> for some <math>M \in \mathbb{R}</math>. But this contradicts the Archimedean property which says that for any <math>M \in \mathbb{R}</math> there exists an <math>n_M \in \mathbb{N}</math> such that <math>M \leq n_M</math>, and so in fact <math>(n)</math> is not bounded and by the divergence criteria, is divergent as well.</p>

==Resources==
* [http://mathonline.wikidot.com/the-divergence-criteria-for-sequences The Divergence Criteria for Sequences], mathonline.wikidot.com

@@ Line 25: / Line 25: @@
 <p>Once again, this sequence is unbounded. Suppose that instead <math>(n)</math> is bounded, that is <math>\mid n \mid = n  M</math> for some <math>M \in \mathbb{R}</math>. But this contradicts the Archimedean property which says that for any <math>M \in \mathbb{R}</math> there exists an <math>n_M \in \mathbb{N}</math> such that <math>M \leq n_M</math>, and so in fact <math>(n)</math> is not bounded and by the divergence criteria, is divergent as well.</p>
-==Resources==
+== Licensing ==
-* [http://mathonline.wikidot.com/the-divergence-criteria-for-sequences The Divergence Criteria for Sequences], mathonline.wikidot.com
+Content obtained and/or adapted from:

@@ Line 3: / Line 3: @@
 <table class="wiki-content-table">
 <tr>
-<td><strong>Theorem 1 (Divergence Criteria for Sequences):</strong> A sequence <math>(a_n)</math> of real numbers is divergent if either one of the following hold:<br />
+<blockquote style="background: white; border: 1px solid black; padding: 1em;">
 <strong>1.</strong> <math>(a_n)</math> has two subsequences <math>(a_{n_k})</math> and <math>(a_{n_p})</math> that converge to two different limits.<br />
 <strong>2.</strong> <math>(a_n)</math> has a subsequence that is divergent.<br />
-<strong>3.</strong> <math>(a_n)</math> is unbounded.</td>
+<strong>3.</strong> <math>(a_n)</math> is unbounded.</blockquote>
 </tr>
 </table>

@@ Line 17: / Line 17: @@
 <p>Since these two subsequences converge to different limits, we conclude that <math>((-1)^n)</math> is divergent.</p>
 <h2 id="toc2">Example 2</h2>
-<p><strong>Show that the sequence <math>(a_n)</math> defined by <math>a_n = \left\{\begin{matrix} 1/n   \text{if n is even} \\ n    \text{if n is odd}\end{matrix}\right.</math>.</strong></p>
+<p><strong>Show that the sequence <math>(a_n)</math> defined by <math>a_n = \left\{\begin{matrix} 1/n  & \text{if n is even} \\ n  &  \text{if n is odd}\end{matrix}\right.</math>.</strong></p>
 <p>Let's first look at a few terms of this sequence. We have that <math>(a_n) = \left(1, \frac{1}{2}, 3, \frac{1}{4}, 5, \frac{1}{6}, ... \right)</math>. We can see this sequence is not bounded above and hence not bounded, which we will prove.</p>
 <p>Suppose that there exists an <math>M \in \mathbb{N}</math> such that <math>\mid a_n \mid  M</math> for all <math>n \in \mathbb{N}</math>. By the Archimedean property since <math>M \in \mathbb{R}</math> there exists a natural number <math>n_M \in \mathbb{N}</math> such that <math>M \leq n_M</math>. We also note that <math>M \leq n_M  n_M + 1</math>. So either <math>n_M</math> or <math>n_M + 1</math> is a term in the sequence <math>(a_n)</math> which contradicts the sequence <math>(a_n)</math> from being bounded.</p>