Sequences and Their Limits - Revision history

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Khanh at 23:51, 21 November 2021

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Khanh at 21:43, 21 November 2021

2021-11-21T21:43:48Z

Khanh: Created page with "Consider the sequence $\left \{ \frac{1}{n} \right \}_{n=1}^{\infty} = \left \{ 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, ... \right \}$ . As $n \to \infty$
2021-11-21T21:34:40Z

Created page with "Consider the sequence <math>\left \{ \frac{1}{n} \right \}_{n=1}^{\infty} = \left \{ 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, ... \right \}</math>. As <math>n \to \infty</mat..."

New page
Consider the sequence <math>\left \{ \frac{1}{n} \right \}_{n=1}^{\infty} = \left \{ 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, ... \right \}</math>. As <math>n \to \infty</math>, it appears as though <math>\frac{1}{n} \to 0</math>. In fact, we know that this is true since <math>\lim_{n \to \infty} \frac{1}{n} = 0</math>. We will now formalize the definition of a limit with regards to sequences.

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
:'''Definition:''' If <math>\{ a_n \}</math> is a sequence, then <math>\lim_{n \to \infty} a_n = L</math> means that for every <math>\epsilon > 0</math> there exists a corresponding <math>N \in \mathbb{N}</math> such that if <math>n \geq N</math>, then <math>\mid a_n - L \mid < \epsilon</math>. If this limit exists, then we say that the sequence <math>\{ a_n \}</math> '''Converges''', and if this limit doesn't exist then we say say the sequence <math>\{ a_n \}</math> '''Diverges'''.
</blockquote>

We note that our definition of the limit of a sequence is very similar to the limit of a function, in fact, we can think of a sequence as a function whose domain is the set of natural numbers <math>\mathbb{N}</math>. From this notion, we obtain the very important theorem:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
:'''Theorem 1:''' If <math>\{ a_n \}</math> is a sequence and a function <math>f(x)</math>, then if <math>f(n) = a_n</math> where <math>n \in \mathbb{N}</math> and <math>\lim_{x \to \infty} f(x) = L</math>, then <math>\lim_{n \to \infty} f(n) = L</math>.
</blockquote>

* '''Proof of Theorem 1:''' Let <math>\epsilon > 0</math> be given. We know that <math>\lim_{x \to \infty} f(x) = L</math> which implies that <math>\forall \epsilon > 0 \exists N \in \mathbb{R}</math> such that if <math>x \geq k</math> then <math>\mid f(x) - L \mid < \epsilon</math>.

* Now we want to show that <math>\forall \epsilon > 0 \exists N \in \mathbb{N}</math> such that if <math>n \geq N</math> then <math>\mid a_n - L \mid < \epsilon</math>. We will choose <math>N = \mathrm{max} \{ 1, \lceil k \rceil \}</math>. This ensures that <math>N</math> is an integer.

* Now since <math>N \geq k</math> then it follows that if <math>\forall n \geq N</math> then <math>\mid f(n) - L \mid < \epsilon</math>. But <math>f(n) = a_n</math> and so <math>\mid a_n - L \mid < \epsilon</math> so <math>\lim_{n \to \infty} a_n = L</math>. <math>\blacksquare</math></li>

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
:'''Important Note:''' The converse of this theorem is not implied to be true! That is if <math>f(n) = a_n</math> and <math>\lim_{n \to \infty} f(n) = L</math>, then this does NOT imply that <math>\lim_{x \to \infty} f(x) = L</math>. For example, consider the sequence <math>\{ \sin (2\pi n) \} = \{ 0, 0, 0, ... \}</math>. Clearly this sequence converges at 0. However, the function <math>f(x) = \sin (2\pi x)</math> does not converge, instead, it diverges as it oscillates between <math>-1</math> and <math>1</math>.
</blockquote>

For example, consider the sequence <math>\left \{ \frac{n + 2}{n} \right \}_{n=1}^{\infty}</math>. If we let <math>f(n) = \frac{n + 2}{n}</math> be a function whose domain is the natural numbers, then we calculate the limit of this function like we have in the past, namely:

<div style="text-align: center;"><math>\begin{align} \lim_{n \to \infty} f(n) = \lim_{n \to \infty} \frac{n + 2}{n} = \lim_{n \to \infty} 1 + \frac{2}{n} = 1 \end{align}</math></div>

Therefore the limit of our sequence <math>f(n) = a_n</math> is 1, that is, <math>\left \{ \frac{n + 2}{n} \right \}_{n=1}^{\infty}</math> converges to 1 as <math>n \to \infty</math>.

Now let's look at another major theorem.

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
:'''Theorem 2:''' If <math>\lim_{n \to \infty} a_n = L</math> and a function <math>f</math> is continuous at <math>L</math>, then <math>\lim_{n \to \infty} f(a_n) = f \left(\lim_{n \to \infty} a_n \right) = f(L)</math>.
</blockquote>

'''Proof of Theorem 2:''' If <math>f</math> is a continuous function at <math>x = L</math>, then we know that <math>f(L)</math> is defined and that <math>\lim_{x \to L} f(x) = f(L)</math>. By the definition of a limit, <math>\forall \epsilon > 0 \exists \delta > 0</math> such that if <math>0 < \mid x - L \mid < \delta</math> then <math>\mid f(x) - f(L) \mid < \epsilon</math>. Let <math>x = a_n</math> so then if <math>0 < \mid a_n - L \mid < \delta</math> then <math>\mid f(a_n) - f(L) \mid < \epsilon</math>.

We want to show that <math>\lim_{n \to \infty} f(a_n) = f(L)</math>, that is <math>\forall \epsilon > 0 \exists N \in \mathbb{N}</math> such that if <math>n \geq N</math> then <math>\mid f(a_n) - f(L) \mid < \epsilon</math>, which is what we showed above.

We will now look at some important limit laws regarding sequences

@@ Line 172: / Line 172: @@
-:*'''Proof of Law 7:''' Let <math>\{ a_n \}</math>, <math>\{ b_n \}</math> and <math>\{ c_n \}</math> be convergent sequences such that <math>a_n \leq b_n \leq c_n</math> is true always after some <math>M^{\mathrm{th}}</math> term, and let <math>\lim_{n \to \infty} a_n = \lim_{n \to \infty} c_n = L</math>. We want to show that <math>\forall \epsilon > 0  \exists N \in \mathbb{N}</math> such that if ><math>n \geq N</math> then ><math>\mid b_n - L \mid < \epsilon</math>.
+:*'''Proof of Law 7:''' Let <math>\{ a_n \}</math>, <math>\{ b_n \}</math> and <math>\{ c_n \}</math> be convergent sequences such that <math>a_n \leq b_n \leq c_n</math> is true always after some <math>M^{\mathrm{th}}</math> term, and let <math>\lim_{n \to \infty} a_n = \lim_{n \to \infty} c_n = L</math>. We want to show that <math>\forall \epsilon > 0  \exists N \in \mathbb{N}</math> such that if <math>n \geq N</math> then ><math>\mid b_n - L \mid < \epsilon</math>.
-:*We note that <math>\lim_{n \to \infty} a_n = L</math> implies that ><math>\forall \epsilon > 0  \exists N_1</math> such that if <math>n \geq N_1</math> then <math>\mid a_n - L \mid < \epsilon</math> or rather, ><math>-\epsilon < a_n - L < \epsilon</math>.
+:*We note that <math>\lim_{n \to \infty} a_n = L</math> implies that <math>\forall \epsilon > 0  \exists N_1</math> such that if <math>n \geq N_1</math> then <math>\mid a_n - L \mid < \epsilon</math> or rather, <math>-\epsilon < a_n - L < \epsilon</math>.
-:*Similarly we note that ><math>\lim_{n \to \infty} c_n = L</math> implies that ><math>\forall \epsilon > 0  \exists N_2</math> such that if <math>n \geq N_2</math> then ><math>\mid c_n - L \mid < \epsilon</math> or rather <math>-\epsilon < c_n - L < \epsilon</math>.
+:*Similarly we note that <math>\lim_{n \to \infty} c_n = L</math> implies that <math>\forall \epsilon > 0  \exists N_2</math> such that if <math>n \geq N_2</math> then <math>\mid c_n - L \mid < \epsilon</math> or rather <math>-\epsilon < c_n - L < \epsilon</math>.
-:*Now let <math>N = \mathrm{max} \{ M, N_1, N_2 \}</math> to ensure that <math>-\epsilon < a_n - L < \epsilon</math>, ><math>-\epsilon < c_n - L \epsilon</math>, and <math>a_n \leq b_n \leq c_n</math>. Subtracting <math>L</math> from all parts of this inequality we get that <math>a_n - L \leq b_n - L \leq c_n - L</math> or rather <math>-\epsilon < b_n - L < \epsilon</math> so then <math>\mid b_n - L \mid < \epsilon</math> and thus <math>\lim_{n \to \infty} b_n = L</math>. ><math>\blacksquare</math>
+:*Now let <math>N = \mathrm{max} \{ M, N_1, N_2 \}</math> to ensure that <math>-\epsilon < a_n - L < \epsilon</math>, <math>-\epsilon < c_n - L \epsilon</math>, and <math>a_n \leq b_n \leq c_n</math>. Subtracting <math>L</math> from all parts of this inequality we get that <math>a_n - L \leq b_n - L \leq c_n - L</math> or rather <math>-\epsilon < b_n - L < \epsilon</math> so then <math>\mid b_n - L \mid < \epsilon</math> and thus <math>\lim_{n \to \infty} b_n = L</math>. ><math>\blacksquare</math>

@@ Line 144: / Line 144: @@
-:*Proof of Law 5:
+:*'''Proof of Law 5:'''
@@ Line 159: / Line 159: @@
 *'''Law 6 (Power Law of Convergent Sequences):''' If the limit of the sequence <math>\{ a_n \}</math> is convergent, that is <math>\lim_{n \to \infty} a_n = A</math> and <math>k</math> is a non-negative integer, then <math>\lim_{n \to \infty} (a_n)^k = \left ( \lim_{n \to \infty} a_n \right )^k = (A)^k</math> provided that <math>k \in \mathbb{N}</math>.
 * '''Law 7 (Squeeze Theorem for Convergent Sequences):''' If the limits of the sequences <math>\{ a_n \}</math>, <math>\{ b_n \}</math>, and <math>\{ c_n \}</math> are convergent and <math>a_n \leq b_n \leq c_n</math> is true always after some <math>n^{\mathrm{th}}</math> term, if <math>\lim_{n \to \infty} a_n = \lim_{n \to \infty} c_n = L</math>, then <math>\lim_{n \to \infty} b_n = L</math>.
@@ Line 172: / Line 192: @@
 Content obtained and/or adapted from:
 * [http://mathonline.wikidot.com/limit-of-a-sequence Limit of a Sequence, mathonline.wikidot.com] under a CC BY-SA license
-* [http://mathonline.wikidot.com/limit-sum-difference-laws-for-convergent-sequences#toc1 Proofs of the Limit Sum/Difference Laws for Convergent Sequences, mathonline.wikidot.com] under a CC BY-SA license
+* [http://mathonline.wikidot.com/limit-sum-difference-laws-for-convergent-sequences#toc1 Limit Sum/Difference Laws for Convergent Sequences, mathonline.wikidot.com] under a CC BY-SA license
 * [http://mathonline.wikidot.com/limit-product-quotient-laws-for-convergent-sequences#toc1 Limit Product/Quotient Laws for Convergent Sequences, mathonline.wikidot.com] under a CC BY-SA license

← Older revision		Revision as of 23:51, 21 November 2021
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	* '''Law 5 (Constant Multiple Law of Convergent Sequences):''' If the limit of the sequence <math>\{ a_n \}</math> is convergent, that is <math>\lim_{n \to \infty} a_n = A</math>, and <math>k</math> is a constant, then <math>\lim_{n \to \infty} ka_n = k \lim_{n \to \infty} a_n = kA</math>.		* '''Law 5 (Constant Multiple Law of Convergent Sequences):''' If the limit of the sequence <math>\{ a_n \}</math> is convergent, that is <math>\lim_{n \to \infty} a_n = A</math>, and <math>k</math> is a constant, then <math>\lim_{n \to \infty} ka_n = k \lim_{n \to \infty} a_n = kA</math>.
		+
		+
		+	:*Proof of Law 5:
		+
		+
		+	:*Let <math>k</math></span> be a constant such that <math>k \neq 0</math>. We note that the case that <math>k = 0</math></span> is trivial since then <math>\lim_{n\ \to \infty} 0a_n = \lim_{n \to \infty} 0 = 0</math>. Let <math>\{ a_n \}</math> be a convergent sequence so that <math>\lim_{n \to \infty} a_n = A</math> which implies that <math>\forall \epsilon \exists N_1 \in \mathbb{N}</math> such that if <math>n \geq N_1</math> then <math>\mid a_n - A \mid < \frac{\epsilon}{\mid k \mid}</math>.
		+
		+
		+	:*Now let <math>\epsilon > 0</math> be given. We want to show <math>\lim_{n \to \infty} ka_n = kA</math>, that is, <math>\forall \epsilon > 0 \exists N \in \mathbb{N}</math> such that if <math>n \geq N</math></span> then <math>\mid k_an - kA \mid < \epsilon</math>. Working with this inequality:
		+
		+	<div style="text-align: center;"><math>\begin{align} \quad \mid k a_n - kA \mid = \mid k(a_n - A) \mid = \mid k \mid \mid a_n - A \mid < \epsilon \end{align}</math></div>
		+
		+
		+	:*We choose <math>N = N_1</math> and therefore if <math>n \geq N</math> then <math>\mid a_n - A \mid < \frac{\epsilon}{\mid k \mid}</math> and so <math>\lim_{n \to \infty} ka_n = kA</math>. <math>\blacksquare</math>

@@ Line 101: / Line 101: @@
 :*And making the appropriate substitutions we get that:
-<div style="text-align: center;"><math>\begin{align} \quad \mid a_nb_n - AB \mid \leq \mid a_n \mid \mid b_n - B \mid + \mid B \mid \mid a_n - A \mid < M \frac{\epsilon}{2 \mathrm{max} \{M, 1\}} + \mid b \mid \frac{\epsilon}{2\mathrm{max} \{ \mid b \mid, 1 \}} &lt; \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}</math></div>
+<div style="text-align: center;"><math>\begin{align} \quad \mid a_nb_n - AB \mid \leq \mid a_n \mid \mid b_n - B \mid + \mid B \mid \mid a_n - A \mid < M \frac{\epsilon}{2 \mathrm{max} \{M, 1\}} + \mid b \mid \frac{\epsilon}{2\mathrm{max} \{ \mid b \mid, 1 \}} < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}</math></div>
@@ Line 159: / Line 159: @@
 * [http://mathonline.wikidot.com/limit-of-a-sequence Limit of a Sequence, mathonline.wikidot.com] under a CC BY-SA license
 * [http://mathonline.wikidot.com/limit-sum-difference-laws-for-convergent-sequences#toc1 Proofs of the Limit Sum/Difference Laws for Convergent Sequences, mathonline.wikidot.com] under a CC BY-SA license

@@ Line 69: / Line 69: @@
 * '''Law 2 (Difference Law of Convergent Sequences):''' If the limits of the sequences <math>\{ a_n \}</math> and <math>\{ b_n \}</math> are convergent, that is <math>\lim_{n \to \infty} a_n = A</math> and <math>\lim_{n \to \infty} b_n = B</math> then <math>\lim_{n \to \infty} (a_n - b_n) = \lim_{n \to \infty} a_n - \lim_{n \to \infty} b_n = A - B</math>.
 *'''Law 3 (Product Law of Convergent Sequences):''' If the limits of the sequences <math>\{ a_n \}</math> and <math>\{ b_n \}</math> are convergent, that is <math>\lim_{n \to \infty} a_n = A</math> and <math>\lim_{n \to \infty} b_n = B</math>, then <math>\lim_{n \to \infty} (a_nb_n) = \lim_{n \to \infty} a_n \cdot \lim_{n \to \infty} b_n = AB</math>.
 *'''Law 4 (Quotient Law of Convergent Sequences):''' If the limits of the sequences <math>\{ a_n \}</math> and <math>\{ b_n \}</math> are convergent, that is <math>\lim_{n \to \infty} a_n = A</math> and <math>\lim_{n \to \infty} b_n = B</math>, then <math>\lim_{n \to \infty} \frac{a_n}{b_n} = \frac{\lim_{n \to \infty} a_n}{\lim_{n \to \infty} b_n} = \frac{A}{B}</math> provided that <math>\lim_{n \to \infty} b_n \neq 0</math>.
@@ Line 94: / Line 158: @@
 Content obtained and/or adapted from:
 * [http://mathonline.wikidot.com/limit-of-a-sequence Limit of a Sequence, mathonline.wikidot.com] under a CC BY-SA license

← Older revision		Revision as of 22:30, 21 November 2021
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	*'''Law 1 (Addition Law of Convergent Sequences):''' If the limits of the sequences <math>\{ a_n \}</math> and <math>\{ b_n \}</math> are convergent, that is <math>\lim_{n \to \infty} a_n = A</math> and <math>\lim_{n \to \infty} b_n = B</math> then <math>\lim_{n \to \infty} (a_n + b_n) = \lim_{n \to \infty} a_n + \lim_{n \to \infty} b_n = A + B</math>.		*'''Law 1 (Addition Law of Convergent Sequences):''' If the limits of the sequences <math>\{ a_n \}</math> and <math>\{ b_n \}</math> are convergent, that is <math>\lim_{n \to \infty} a_n = A</math> and <math>\lim_{n \to \infty} b_n = B</math> then <math>\lim_{n \to \infty} (a_n + b_n) = \lim_{n \to \infty} a_n + \lim_{n \to \infty} b_n = A + B</math>.
		+
		+
		+	:*'''Proof of Law 1:''' Let <math>\{ a_n \}</math> and <math>\{ b_n \}</math> be convergent sequences. By the definition of a sequence being convergent, we know that <math>\lim_{n \to \infty} a_n = A</math> and <math>\lim_{n \to \infty} b_n = B</math> for some <math>A, B \in \mathbb{R}</math>.
		+
		+
		+	:*Now let <math>\epsilon > 0</math> be given, and recall that <math>\lim_{n \to \infty} a_n = A</math> implies that <math>\forall \epsilon \exists N_1 \in \mathbb{N}</math> such that if <math>n \geq N_1</math> then <math>\mid a_n - A \mid < \frac{\epsilon}{2}</math>. Similarly, <math>\lim_{n \to \infty} b_n = B</math> implies that <math>\forall \epsilon \exists N_2 \in \mathbb{N}</math> such that if <math>n \geq N_2</math> then <math>\mid b_n - B \mid < \frac{\epsilon}{2}</math>.
		+
		+
		+	:*We will choose <math>N = \mathrm{max} \{ N_1, N_2 \}</math>. By choosing the larger of <math>N_1</math> and <math>N_2</math>, we ensure that the if <math>n \geq \mathrm{max} \{ N_1, N_2 \}</math>, then <math>\mid a_n - A \mid < \frac{\epsilon}{2}</math> and <math>\mid b_n - B \mid < \frac{\epsilon}{2}</math>. We now want to show that if <math>n \geq N</math>, then <math>\mid (a_n + b_n) - (A + B) \mid < \epsilon</math>. By the triangle inequality we obtain that:
		+
		+
		+	<div style="text-align: center;"><math>\begin{align} \quad \quad \mid (a_n + b_n) - (A + B) \mid = \mid (a_n - A) + (b_n - B) \mid \leq \mid a_n + A \mid + \mid b_n + B \mid < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}</math></div>
		+
		+
		+	:*Therefore <math>\lim_{n \to \infty} (a_n + b_n) = A + B</math>. <math>\blacksquare</math>
		+


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	Let <math>f(n) = \frac{n^2 - 1}{n + 1}</math> be a function analogous to our sequence. When we factor the numerator and cancel like-terms, we get that <math>\lim_{n \to \infty} f(n) = \lim_{n \to \infty} \frac{n^2 - 1}{n + 1} = \lim_{n \to \infty} \frac{(n +1)(n - 1)}{n + 1} = \lim_{n \to \infty} n - 1 = \infty</math>. Therefore the sequence <math>\left \{ \frac{x^2 - 1}{x + 1} \right \}_{n=1}^{\infty}</math> is divergent.		Let <math>f(n) = \frac{n^2 - 1}{n + 1}</math> be a function analogous to our sequence. When we factor the numerator and cancel like-terms, we get that <math>\lim_{n \to \infty} f(n) = \lim_{n \to \infty} \frac{n^2 - 1}{n + 1} = \lim_{n \to \infty} \frac{(n +1)(n - 1)}{n + 1} = \lim_{n \to \infty} n - 1 = \infty</math>. Therefore the sequence <math>\left \{ \frac{x^2 - 1}{x + 1} \right \}_{n=1}^{\infty}</math> is divergent.
		+
		+	== Licensing ==
		+	Content obtained and/or adapted from:
		+	* [http://mathonline.wikidot.com/limit-of-a-sequence Limit of a Sequence, mathonline.wikidot.com] under a CC BY-SA license

← Older revision		Revision as of 21:43, 21 November 2021
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	We will now look at some important limit laws regarding sequences		We will now look at some important limit laws regarding sequences
		+
		+	==Limit Laws of Convergent Sequences==
		+
		+	We will now look at some very important limit laws regarding convergent sequences, all of which are analogous to the limit laws for functions that we already know of.
		+
		+
		+	*'''Law 1 (Addition Law of Convergent Sequences):''' If the limits of the sequences <math>\{ a_n \}</math> and <math>\{ b_n \}</math> are convergent, that is <math>\lim_{n \to \infty} a_n = A</math> and <math>\lim_{n \to \infty} b_n = B</math> then <math>\lim_{n \to \infty} (a_n + b_n) = \lim_{n \to \infty} a_n + \lim_{n \to \infty} b_n = A + B</math>.
		+
		+
		+	* '''Law 2 (Difference Law of Convergent Sequences):''' If the limits of the sequences <math>\{ a_n \}</math> and <math>\{ b_n \}</math> are convergent, that is <math>\lim_{n \to \infty} a_n = A</math> and <math>\lim_{n \to \infty} b_n = B</math> then <math>\lim_{n \to \infty} (a_n - b_n) = \lim_{n \to \infty} a_n - \lim_{n \to \infty} b_n = A - B</math>.
		+
		+
		+	*'''Law 3 (Product Law of Convergent Sequences):''' If the limits of the sequences <math>\{ a_n \}</math> and <math>\{ b_n \}</math> are convergent, that is <math>\lim_{n \to \infty} a_n = A</math> and <math>\lim_{n \to \infty} b_n = B</math>, then <math>\lim_{n \to \infty} (a_nb_n) = \lim_{n \to \infty} a_n \cdot \lim_{n \to \infty} b_n = AB</math>.
		+
		+
		+	*'''Law 4 (Quotient Law of Convergent Sequences):''' If the limits of the sequences <math>\{ a_n \}</math> and <math>\{ b_n \}</math> are convergent, that is <math>\lim_{n \to \infty} a_n = A</math> and <math>\lim_{n \to \infty} b_n = B</math>, then <math>\lim_{n \to \infty} \frac{a_n}{b_n} = \frac{\lim_{n \to \infty} a_n}{\lim_{n \to \infty} b_n} = \frac{A}{B}</math> provided that <math>\lim_{n \to \infty} b_n \neq 0</math>.
		+
		+
		+	* '''Law 5 (Constant Multiple Law of Convergent Sequences):''' If the limit of the sequence <math>\{ a_n \}</math> is convergent, that is <math>\lim_{n \to \infty} a_n = A</math>, and <math>k</math> is a constant, then <math>\lim_{n \to \infty} ka_n = k \lim_{n \to \infty} a_n = kA</math>.
		+
		+
		+	*'''Law 6 (Power Law of Convergent Sequences):''' If the limit of the sequence <math>\{ a_n \}</math> is convergent, that is <math>\lim_{n \to \infty} a_n = A</math> and <math>k</math> is a non-negative integer, then <math>\lim_{n \to \infty} (a_n)^k = \left ( \lim_{n \to \infty} a_n \right )^k = (A)^k</math> provided that <math>k \in \mathbb{N}</math>.
		+
		+
		+	* '''Law 7 (Squeeze Theorem for Convergent Sequences):''' If the limits of the sequences <math>\{ a_n \}</math>, <math>\{ b_n \}</math>, and <math>\{ c_n \}</math> are convergent and <math>a_n \leq b_n \leq c_n</math> is true always after some <math>n^{\mathrm{th}}</math> term, if <math>\lim_{n \to \infty} a_n = \lim_{n \to \infty} c_n = L</math>, then <math>\lim_{n \to \infty} b_n = L</math>.
		+
		+
		+	==Example 1==
		+	'''Determine whether the sequence <math>\left \{ \frac{x^2 - 1}{x + 1} \right \}_{n=1}^{\infty}</math> is convergent or divergent.'''
		+
		+	Let <math>f(n) = \frac{n^2 - 1}{n + 1}</math> be a function analogous to our sequence. When we factor the numerator and cancel like-terms, we get that <math>\lim_{n \to \infty} f(n) = \lim_{n \to \infty} \frac{n^2 - 1}{n + 1} = \lim_{n \to \infty} \frac{(n +1)(n - 1)}{n + 1} = \lim_{n \to \infty} n - 1 = \infty</math>. Therefore the sequence <math>\left \{ \frac{x^2 - 1}{x + 1} \right \}_{n=1}^{\infty}</math> is divergent.