Properly Divergent Sequences - Revision history

Khanh at 21:48, 7 November 2021

2021-11-07T21:48:56Z

Khanh at 21:47, 7 November 2021

2021-11-07T21:47:57Z

Khanh at 21:46, 7 November 2021

2021-11-07T21:46:08Z

Khanh: Created page with "Recall that a sequence $(a_n)$ of real numbers is said to be convergent to the real number $A$ if $\forall \epsilon > 0$ ther..."

2021-11-07T21:21:00Z

Created page with "Recall that a sequence <math>(a_n)</math> of real numbers is said to be <strong>convergent</strong> to the real number <math>A</math> if <math>\forall \epsilon > 0</math> ther..."

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Recall that a sequence <math>(a_n)</math> of real numbers is said to be <strong>convergent</strong> to the real number <math>A</math> if <math>\forall \epsilon > 0</math> there exists an <math>N \in \mathbb{N}</math> such that if <math>n \geq N</math> then <math>\mid a_n - A \mid < \epsilon</math>.

If we negate this statement we have that a sequence <math>(a_n)</math> of real numbers is '''divergent''' if <math>\forall A \in \mathbb{R}</math> then <math>\exists \epsilon_0 > 0</math> such that <math>\forall N \in \mathbb{N}</math> such that if <math>n \geq N</math> then <math>\mid a_n - A \mid \geq \epsilon_0</math>. However, there are different types of divergent sequences. For example, a sequence can alternate between different points and be divergent such as the sequence <math>((-1)^n)</math>, or instead, the sequence can tend to infinity such as <math>(n)</math> or negative infinity such as <math>(-n)</math>, or neither, such as <math>((-1)^n(n))</math>. We will now define '''properly divergent sequences'''.

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
'''Definition:''' A sequence of real numbers <math>(a_n)</math> is said to be '''Properly Divergent to <math>\infty</math>''' if <math>\displaystyle{\lim_{n \to \infty} a_n = \infty}</math>, that is <math>\forall M \in \mathbb{R}</math> there exists an <math>N \in \mathbb{N}</math> such that if <math>n \geq N</math> then <math>a_n > M</math>. Similarly, <math>(a_n)</math> is said to be '''Properly Divergent to <math>-\infty</math>''' if <math>\displaystyle{\lim_{n \to \infty} a_n = -\infty}</math>, that is <math>\forall M \in \mathbb{R}</math> there exists an <math>N \in \mathbb{N}</math> such that if <math>n \geq N</math> then <math>a_n < M</math>.
</blockquote>

Now let's look at some theorems regarding properly divergent sequences.

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
'''Theorem 1:''' An increasing sequence of real numbers <math>(a_n)</math> is properly divergent to <math>\infty</math> if it is unbounded. A decreasing sequence of real numbers <math>(a_n)</math> is properly divergent to <math>-\infty</math> if it is unbounded.
</blockquote>

*'''Proof:''' Suppose that <math>(a_n)</math> is a sequence of real numbers that is increasing. Since <math>(a_n)</math> is unbounded, then for any <math>M \in \mathbb{R}</math> there exists a term <math>a_M</math> (dependent on <math>M</math>) such that <math>M < a_M</math>. Since <math>(a_n)</math> is an increasing sequence, then for <math>n \geq M</math> we have that <math>M < a_n</math> and since <math>M</math> is arbitrary we have that <math>\lim_{n \to \infty} a_n = \infty</math>.
*Similarly suppose that <math>(a_n)</math> is a sequence of real numbers that is decreasing. Since <math>(a_n)</math> is unbounded, then for any <math>M \in \mathbb{R}</math> there exists a term <math>a_M</math> (dependent on <math>M</math> such that <math>M > a_M</math>. Since <math>(a_n)</math> is a decreasing sequence, then for <math>n \geq M</math> we have that <math>M > a_n</math> and since <math>M</math> is arbitrary we have that <math>\lim_{n \to \infty} a_n = -\infty</math>. <math>\blacksquare</math>

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
'''Theorem 2:''' Let <math>(a_n)</math> and <math>(b_n)</math> be sequences of real numbers such that <math>a_n \leq b_n</math> for all <math>n \in \mathbb{N}</math>. Then if <math>\lim_{n \to \infty} a_n = \infty</math> then <math>\lim_{n \to \infty} b_n = \infty</math>.
</blockquote>

*'''Proof:''' Let <math>(a_n)</math> and <math>(b_n)</math> be sequences of real numbers such that <math>a_n \leq b_n</math> for all <math>n \in \mathbb{N}</math>, and let <math>\lim_{n \to \infty} a_n = \infty</math>. Then it follows that for all <math>M \in \mathbb{R}</math> that there exists an <math>N</math> (dependent on <math>M</math> such that if <math>n \geq N</math> then <math>a_n \geq M</math>. But we have that <math>b_n \geq a_n</math> for all <math>n \in \mathbb{N}</math> and so for <math>n \geq N</math> we have that <math>b_n \geq M</math>. Since <math>M</math> is arbitrary it follows that <math>\lim_{n \to \infty} b_n = \infty</math>. <math>\blacksquare</math>

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
'''Theorem 3:''' Let <math>(a_n)</math> and <math>(b_n)</math> be sequences of real numbers such that <math>a_n \leq b_n</math> for all <math>n \in \mathbb{N}</math>. Then if <math>\lim_{n \to \infty} b_n = -\infty</math> then <math>\lim_{n \to \infty} a_n = \infty</math>.
</blockquote>

*'''Proof:''' Let <math>(a_n)</math> and <math>(b_n)</math> be sequences of real numbers such that <math>a_n \leq b_n</math> for all <math>n \in \mathbb{N}</math>, and let <math>\lim_{n \to \infty} b_n = -\infty</math>. Then it follows that for all <math>M \in \mathbb{R}</math> that there exists an <math>N</math> (dependent on <math>M</math> such that if <math>n \geq N</math> then <math>b_n \leq M</math>. But we have that <math>a_n \leq b_n</math> for all <math>n \in \mathbb{N}</math> and so for <math>n \geq N</math> we have that <math>a_n \leq M</math>. Since <math>M</math> is arbitrary it follows that <math>\lim_{n \to \infty} a_n = -\infty</math>. <math>\blacksquare</math>

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
'''Theorem 4:''' If <math>(a_n)</math> and <math>(b_n)</math> are sequences of positive real numbers suppose that for some real number <math>L > 0</math> that <math>\lim_{n \to \infty} \frac{a_n}{b_n} = L</math>. Then <math>\lim_{n \to \infty} a_n = \infty</math> if and only if <math>\lim_{n \to \infty} b_n = \infty</math>.
</blockquote>

*'''Proof:''' Suppose that <math>(a_n)</math> and <math>(b_n)</math> are convergent sequences and that <math>\lim_{n \to \infty} \frac{a_n}{b_n} = L</math> for <math>L \in \mathbb{R}</math> and <math>L > 0</math>. Then for <math>\epsilon = \frac{L}{2} > 0</math> we have that for some <math>N \in \mathbb{N}</math> if <math>n \geq N</math> then <math>\mid \frac{a_n}{b_n} - L \mid < \frac{L}{2}</math> or equivalently:

← Older revision		Revision as of 21:48, 7 November 2021
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−	~~==Properly Divergent Sequences==~~	+
	Recall that a sequence <math>(a_n)</math> of real numbers is said to be <strong>convergent</strong> to the real number <math>A</math> if <math>\forall \epsilon > 0</math> there exists an <math>N \in \mathbb{N}</math> such that if <math>n \geq N</math> then <math>\mid a_n - A \mid < \epsilon</math>.		Recall that a sequence <math>(a_n)</math> of real numbers is said to be <strong>convergent</strong> to the real number <math>A</math> if <math>\forall \epsilon > 0</math> there exists an <math>N \in \mathbb{N}</math> such that if <math>n \geq N</math> then <math>\mid a_n - A \mid < \epsilon</math>.

← Older revision		Revision as of 21:47, 7 November 2021
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		+	==Properly Divergent Sequences==
	Recall that a sequence <math>(a_n)</math> of real numbers is said to be <strong>convergent</strong> to the real number <math>A</math> if <math>\forall \epsilon > 0</math> there exists an <math>N \in \mathbb{N}</math> such that if <math>n \geq N</math> then <math>\mid a_n - A \mid < \epsilon</math>.		Recall that a sequence <math>(a_n)</math> of real numbers is said to be <strong>convergent</strong> to the real number <math>A</math> if <math>\forall \epsilon > 0</math> there exists an <math>N \in \mathbb{N}</math> such that if <math>n \geq N</math> then <math>\mid a_n - A \mid < \epsilon</math>.

← Older revision		Revision as of 21:46, 7 November 2021
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	*'''Proof:''' Suppose that <math>(a_n)</math> and <math>(b_n)</math> are convergent sequences and that <math>\lim_{n \to \infty} \frac{a_n}{b_n} = L</math> for <math>L \in \mathbb{R}</math> and <math>L > 0</math>. Then for <math>\epsilon = \frac{L}{2} > 0</math> we have that for some <math>N \in \mathbb{N}</math> if <math>n \geq N</math> then <math>\mid \frac{a_n}{b_n} - L \mid < \frac{L}{2}</math> or equivalently:		*'''Proof:''' Suppose that <math>(a_n)</math> and <math>(b_n)</math> are convergent sequences and that <math>\lim_{n \to \infty} \frac{a_n}{b_n} = L</math> for <math>L \in \mathbb{R}</math> and <math>L > 0</math>. Then for <math>\epsilon = \frac{L}{2} > 0</math> we have that for some <math>N \in \mathbb{N}</math> if <math>n \geq N</math> then <math>\mid \frac{a_n}{b_n} - L \mid < \frac{L}{2}</math> or equivalently:
		+	<div style="text-align: center;"> <math> \begin{align} \frac{L}{2} < \frac{a_n}{b_n} < \frac{3L}{2} \\ \frac{L}{2}b_n < a_n < \frac{3L}{2}b_n \\ \end{align}</math></div>
		+
		+	If <math>\lim_{n \to \infty} a_n = \infty</math> then since <math>a_n < \frac{3L}{2} b_n</math> it follows that <math>\lim_{n \to \infty} b_n = \infty</math>. Similarly if <math>\lim_{n \to \infty} b_n = \infty</math> then since <math>\frac{L}{2} b_n < a_n</math> it follows that <math>\lim_{n \to \infty} a_n = \infty</math>. <math>\blacksquare</math>
		+
		+	<blockquote style="background: white; border: 1px solid black; padding: 1em;">
		+	'''Theorem 5:''' If <math>(a_n)</math> is a properly divergent subsequence then there exists no convergent subsequences <math>(a_{n_k})</math> of <math>(a_n)</math>.
		+	</blockquote>
		+
		+	* '''Proof:''' We will first deal with the case where <math>(a_n)</math> is properly divergent to <math>\infty</math>. Suppose instead that there exists a subsequence <math>(a_{n_k})</math> that converges to <math>L</math>. Then <math>\forall \epsilon > 0</math> <math>\exists K \in \mathbb{N}</math> such that if <math>k \geq K</math> then <math>\mid a_{n_k} - L \mid < \epsilon</math>, and so for <math>k \geq K</math> then <math>L - \epsilon < a_{n_k} < L + \epsilon</math>, and so <math>a_{n_k} < L + \epsilon</math>.
		+
		+	* Now if <math>(a_n)</math> diverges to <math>\infty</math> then for <math>L + \epsilon \in \mathbb{R}</math> <math>\exists N \in \mathbb{N}</math> such that if <math>n \geq N</math> then <math>a_n > L + \epsilon</math>. So for <math>n_k \geq \mathrm{max} \{ K, N \}</math>, we have that <math>L + \epsilon < a_{n_k} < L + \epsilon</math> which is a contradiction. So our assumption that <math>(a_{n_k})</math> converges was false, and so there exists no convergent subsequences <math>(a_{n_k})</math>. <math>\blacksquare</math>
		+
		+	===Example 1===
		+	'''Show that the sequence <math>(n^2)</math> is properly divergent to <math>\infty</math>.'''
		+
		+	We want to show that <math>\forall M \in \mathbb{R}</math> there exists an <math>N \in \mathbb{N}</math> such that if <math>n \geq N</math> then <math>n^2 > M</math>. Notice that <math>n^2 > n</math> for all <math>n \in \mathbb{N}</math>. By the Archimedean property, since <math>M \in \mathbb{R}</math> there exists an <math>n \in \mathbb{N}</math> such that <math>M \leq n</math>, and so <math>n^2 > n \geq M</math>. Therefore the sequence <math>(n^2)</math> diverges properly to <math>\infty</math>.
		+
		+	== Licensing ==
		+	Content obtained and/or adapted from:
		+	* [http://mathonline.wikidot.com/properly-divergent-sequences Properly Divergent Sequences, mathonline.wikidot.com] under a CC BY-SA license

Properly Divergent Sequences - Revision history

Khanh at 21:48, 7 November 2021

Khanh at 21:47, 7 November 2021

Khanh at 21:46, 7 November 2021

Khanh: Created page with "Recall that a sequence (a_n) of real numbers is said to be convergent to the real number A if \forall \epsilon > 0 ther..."

Khanh: Created page with "Recall that a sequence $(a_n)$ of real numbers is said to be convergent to the real number $A$ if $\forall \epsilon > 0$ ther..."