Skills Review for Alternating Series and Ratio and Root Tests

Learning Outcomes

Apply factorial notation
Simplify expressions using the Product Property of Exponents
Calculate the limit of a function as 𝑥 increases or decreases without bound
Recognize when to apply L’Hôpital’s rule

In the Alternating Series and Ratio and Root Tests sections, we will learn about the last few methods that can be used to determine whether an infinite series diverges or converges. Here we will review how to use factorial notation, use product rule for exponents, take limits at infinity, and apply L’Hopital’s Rule.

Apply Factorial Notation

Recall that $n$ factorial, written as $n!$ , is the product of the positive integers from 1 to $n$ . For example,

\begin{aligned} 4! & = 4 \cdot 3 \cdot 2 \cdot 1 = 24 \\ 5! & = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120 \end{aligned}

$\begin{align}4!&=4\cdot 3\cdot 2\cdot 1=24 \\ 5!&=5\cdot 4\cdot 3\cdot 2\cdot 1=120\\ \text{ } \end{align}$

An example of formula containing a factorial is ${a}_{n}=\left(n+1\right)!$ . The sixth term of the sequence can be found by substituting 6 for $n$ .

\begin{aligned} a_{6} = (6 + 1)! = 7! = 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 5040 \end{aligned}

$\begin{align}{a}_{6}=\left(6+1\right)!=7!=7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1=5040 \\ \text{ }\end{align}$

The factorial of any whole number $n$ is $n\left(n - 1\right)!$ We can therefore also think of $5!$ as $5\cdot 4!\text{.}$

A GENERAL NOTE: FACTORIAL

n factorial is a mathematical operation that can be defined using a recursive formula. The factorial of $n$ , denoted $n!$ , is defined for a positive integer $n$ as:

\begin{array}{l} 0! = 1 \\ 1! = 1 \\ n! = n (n - 1) (n - 2) \dots (2) (1), for n \geq 2 \end{array}

$\begin{array}{l}0!=1\\ 1!=1\\ n!=n\left(n - 1\right)\left(n - 2\right)\cdots \left(2\right)\left(1\right)\text{, for }n\ge 2\end{array}$

The special case $0!$ is defined as $0!=1$ .

Try It

Expand $(n+3)!$ .

Show Solution

$(n+3)!=(n+3)(n+2)(n+1)n(n-1)\cdot...\cdot1$

Notice also that $(n+3)!=(n+3)(n+2)(n+1)n!$

Use the Product Rule for Exponents

A General Note: The Product Rule of Exponents

For any real number $a$ and natural numbers $m$ and $n$ , the product rule of exponents states that

a^{m} \cdot a^{n} = a^{m + n}

${a}^{m}\cdot {a}^{n}={a}^{m+n}$

Example: Using the Product Rule

Write each of the following products with a single base. Do not simplify further.

${t}^{5}\cdot {t}^{3}$
$\left(-3\right)^{5}\cdot \left(-3\right)$
${x}^{2}\cdot {x}^{5}\cdot {x}^{3}$

Show Solution

Use the product rule to simplify each expression.

${t}^{5}\cdot {t}^{3}={t}^{5+3}={t}^{8}$
${\left(-3\right)}^{5}\cdot \left(-3\right)={\left(-3\right)}^{5}\cdot {\left(-3\right)}^{1}={\left(-3\right)}^{5+1}={\left(-3\right)}^{6}$
${x}^{2}\cdot {x}^{5}\cdot {x}^{3}$

At first, it may appear that we cannot simplify a product of three factors. However, using the associative property of multiplication, begin by simplifying the first two.

x^{2} \cdot x^{5} \cdot x^{3} = (x^{2} \cdot x^{5}) \cdot x^{3} = (x^{2 + 5}) \cdot x^{3} = x^{7} \cdot x^{3} = x^{7 + 3} = x^{10}

${x}^{2}\cdot {x}^{5}\cdot {x}^{3}=\left({x}^{2}\cdot {x}^{5}\right)\cdot {x}^{3}=\left({x}^{2+5}\right)\cdot {x}^{3}={x}^{7}\cdot {x}^{3}={x}^{7+3}={x}^{10}$

Notice we get the same result by adding the three exponents in one step.

x^{2} \cdot x^{5} \cdot x^{3} = x^{2 + 5 + 3} = x^{10}

${x}^{2}\cdot {x}^{5}\cdot {x}^{3}={x}^{2+5+3}={x}^{10}$

Try It

RECALL

For any real number $a$ and positive integers $m$ and $n$ , the power rule of exponents states that

{(a^{m})}^{n} = a^{m \cdot n}

${\left({a}^{m}\right)}^{n}={a}^{m\cdot n}$

For an expression like $(a^2)^3$ , you have a base of $a$ raised to the power of $2$ , which is then raised to another power of $3$ . Multiply the exponents $2$ and $3$ to find the new exponent for $a$ . This gives you $a^{2\cdot3}$ or $a^6$ . Always remember: when an exponent is raised to another exponent, multiply the exponents to simplify the expression.

Try It

Simplify the expression $(3a^2b)^3 \cdot (2ab^4)$ .

Show Solution

$(xy)^n=x^n \cdot y^n$ , so we apply the exponent $3$ to both $3a^2$ and $b$ within the parentheses.

(3 a^{2} b)^{3} = 3^{3} \cdot (a^{2})^{3} \cdot b^{3}

$(3a^2b)^3 = 3^3 \cdot (a^2)^3 \cdot b^3$

Now we’ll compute the powers individually.

\begin{array}{l} 3^{3} = 27 \\ (a^{2})^{3} = a^{2 \cdot 3} = a^{6} \\ b^{3} = b^{3} (no change since it's already in exponent form) \end{array}

$\begin{array}{l} 3^3 = 27 \\ (a^2)^3 = a^{2\cdot3} = a^6 \\ b^3 = b^3 \text{ (no change since it's already in exponent form)} \end{array}$

Put together the results of the individual powers.

(3 a^{2} b)^{3} = 27 \cdot a^{6} \cdot b^{3}

$(3a^2b)^3=27 \cdot a^6 \cdot b^3$

Now we multiply the result by the second term $2ab^4$ . Since we are dealing with multiplication, we can multiply coefficients (the numerical parts) and variables with the same base separately.

(27 \cdot a^{6} \cdot b^{3}) \cdot (2 a b^{4}) = (27 \cdot 2) \cdot (a^{6} \cdot a) \cdot (b^{3} \cdot b^{4})

$(27⋅a^6 \cdot b^3) \cdot (2ab^4) = (27 \cdot 2) \cdot (a^6 \cdot a) \cdot (b^3 \cdot b^4)$

Multiply the coefficients and add the exponents for the variables with the same base.

\begin{array}{l} 27 \cdot 2 = 54 \\ a^{6} \cdot a = a^{6 + 1} = a^{7} \\ b^{3} \cdot b^{4} = b^{3 + 4} = b^{7} \end{array}

$\begin{array}{l} 27 \cdot 2 = 54 \\ a^6 \cdot a = a^{6+1} = a^7 \\ b^3 \cdot b^4 = b^{3+4} = b^7 \end{array}$

Finally, combine all the simplified terms to get the final answer.

54 \cdot a^{7} \cdot b^{7}

$54 \cdot a^7 \cdot b^7$

So, the expression $(3a^2b)^3 \cdot (2ab^4)$ simplifies to $54a^7b^7$ .

Try It

Simplify the expression $(2y^2)^3 \cdot (4y^5).$

Show Solution

First, apply the power to the first term, $(2y^2)^3$ , by multiplying the exponent outside the parenthesis with each exponent inside the parenthesis:

(2 y^{2})^{3} = 2^{3} \cdot (y^{2})^{3} = 8 y^{6}

$(2y^2)^3=2^3 \cdot (y^2)^3=8y^6$

Now, multiply this result by $4y^5$ :

8 y^{6} \cdot 4 y^{5} = (8 \cdot 4) \cdot y^{6 + 5} = 32 y^{11}

$8y^6 \cdot 4y^5 = (8 \cdot 4) \cdot y^{6+5} = 32y^{11}$

Therefore, the expression simplifies to $32y^{11}$ .

Take Limits at Infinity

(see Module 5, Skills Review for Sequences.)

Infinite Limits at Infinity

(see Module 5, Skills Review for Sequences.)

Apply L’Hôpital’s Rule

(see Module 5, Skills Review for Sequences.)

Sequences and Series: Prerequisite Material