Sequences Defined by a Recursive Formula

The formulas for some sequences include products of consecutive positive integers. $n$ factorial, written as $n!$, is the product of the positive integers from 1 to $n$. For example,

$\begin{array}{ll} 4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24 \\ 5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120 \end{array}$

$a_6=(6+1)!=7!=7⋅6⋅5⋅4⋅3⋅2⋅1=5040$

The factorial of any whole number $n$ is $n(n−1)!$ We can therefore also think of $5!$ as $5⋅4!$.

$n$ 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:

$\begin{array}{ll} 0! = 1 \\ 1! = 1 \\ n!=n(n-1)(n-2) \cdot\cdot\cdot(2)(1), \text { for } n \geq 2 \end{array}$

Q&A

Can factorials always be found using a calculator?

No. Factorials get large very quickly – faster than even exponential functions! When the output gets too large for the calculator, it will not be able to calculate the factorial.

Example 7

Writing the Terms of a Sequence Using Factorials

Write the first five terms of the sequence defined by the explicit formula $a_n=\frac{5n}{(n+2)!}$.

Solution

Substitute $n=1,n=2$, and so on in the formula.

$\begin{array}{ll}n=1 & a_{1}=\frac{5(1)}{(1+2) !}=\frac{5}{3 !}=\frac{5}{3 \cdot 2 \cdot 1}=\frac{5}{6} \\n=2 & a_{2}=\frac{5(2)}{(2+2) !}=\frac{10}{4 !}=\frac{10}{4 \cdot 3 \cdot 2 \cdot 1}=\frac{5}{12} \\n=3 & a_{3}=\frac{5(3)}{(3+2) !}=\frac{15}{5 !}=\frac{15}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}=\frac{1}{8} \\n=4 & a_{4}=\frac{5(4)}{(4+2) !}=\frac{20}{6 !}=\frac{20}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}=\frac{1}{36} \\n=5 & a_{5}=\frac{5(5)}{(5+2) !}=\frac{25}{7 !}=\frac{25}{7 \cdot 6 \cdot 5 \cdot 4 \cdot \cdot \cdot \cdot \cdot \cdot 1}=\frac{5}{1,008}\end{array}$

The first five terms are $\{\frac{5}{6},\frac{5}{12},\frac{1}{8},\frac{1}{36},\frac{5}{1,008}\}$.

Analysis

Figure 7 shows the graph of the sequence. Notice that, since factorials grow very quickly, the presence of the factorial term in the denominator results in the denominator becoming much larger than the numerator as $n$ increases. This means the quotient gets smaller and, as the plot of the terms shows, the terms are decreasing and nearing zero.

Figure 7

Try It #9

Write the first five terms of the sequence defined by the explicit formula $a_n=\frac{(n+1)!}{2n}$.