Introduction to Sequences and Factorials, precalculus lesson

A thermometer read once every hour, the balance of a savings account at the end of each year, the number of tiles in each row of a staircase pattern. Each gives a first number, a second, a third and so on, with nothing in between. Such an ordered list is a sequence, and it is really a function whose inputs are positions.

Key terms

Sequence as a function of an integer index

Definition A sequence is a function whose domain is a set of consecutive integers, either $\{n_0, n_0 + 1, \dots, N\}$ for integers $n_0 \le N$ or $\{n_0, n_0 + 1, n_0 + 2, \dots\}$ for an integer $n_0$ . In this course its values are real numbers. The value at the integer $n$ is written $a_n$ instead of $a(n)$ and is called the term with index $n$ . The whole sequence is written $\{a_n\}$ , or $\{a_n\}_{n=1}^{\infty}$ when the indices should be shown.

In plain words A sequence is a list of numbers in a definite order, with a position number attached to each entry. The position is the input and the entry is the output.

Example The sequence $a_n = \sqrt{n}$ for $n \ge 1$ starts like this.

$n$	$1$	$2$	$3$	$4$	$5$
$a_n$	$1$	$\sqrt{2} \approx 1.414$	$\sqrt{3} \approx 1.732$	$2$	$\sqrt{5} \approx 2.236$

Its graph is the separate dots $(1, 1)$ , $(2, \sqrt{2})$ , $(3, \sqrt{3})$ and so on. They sit on the curve $y = \sqrt{x}$ , yet the sequence has no value at $n = 2.5$ , because $2.5$ is not an index. The curve is one unbroken piece, while the sequence is a row of isolated points.

Indexing convention

Definition The indexing convention of a sequence is the choice of its first index $n_0$ . Most texts start at $n_0 = 1$ , so that $a_1$ is the first term and $a_n$ is the $n$ th term. Starting at $n_0 = 0$ is common when $a_0$ stands for a starting amount. With first index $n_0$ the $k$ th term is $a_{n_0 + k - 1}$ , so the same list of numbers needs a different formula when the first index changes.

In plain words Before you use a formula, check where the counting starts. Starting at $0$ instead of $1$ shifts every position by one.

Example The odd numbers $1, 3, 5, 7, \dots$ can be written as $a_n = 2n - 1$ for $n \ge 1$ or as $b_n = 2n + 1$ for $n \ge 0$ . The tenth odd number is $a_{10} = 19$ in the first description and $b_9 = 19$ in the second. Writing $b_{10}$ would give $21$ , which is the eleventh odd number.

Explicit term

Definition An explicit formula for a sequence gives the term $a_n$ directly as an expression in the index $n$ , valid for every index in the domain. This expression is called the general term or the $n$ th term. To find a particular term, substitute its index. Finitely many terms never determine an infinite sequence on their own, so an infinite sequence is known only once its rule is given.

In plain words An explicit formula lets you jump straight to any position, the hundredth term as easily as the second. A pattern you spot in the first few terms is a guess until a rule backs it up.

Example An account starts with 1000 dollars and earns 5 percent interest once a year. Its balance after $n$ years is $a_n = 1000(1.05)^n$ for $n \ge 0$ , where $a_0 = 1000$ is the starting balance. After three years

a_3 = 1000 \cdot 1.05^3 = 1157.625,

about 1157.63 dollars. Guessing from a few terms is risky. With $n \ge 1$ the numbers $2, 4, 8$ are the first terms of $a_n = 2^n$ , which continues with $16$ . They are also the first terms of $b_n = n^2 - n + 2$ , which continues with $14$ .

Finite and infinite sequences

Definition A sequence is finite when its domain is a finite set $\{n_0, n_0 + 1, \dots, N\}$ . It then has $N - n_0 + 1$ terms and a last term $a_N$ . A sequence is infinite when its domain is $\{n_0, n_0 + 1, n_0 + 2, \dots\}$ , so it has a term for every integer from $n_0$ on and no last term. Calculus works only with infinite sequences, and there the word sequence on its own always means an infinite one.

In plain words A finite sequence stops. An infinite sequence goes on forever, which is what the three dots at the end of a list mean.

Example The numbers of days in the months of a year that is not a leap year form the finite sequence

\begin{gathered} 31, 28, 31, 30, 31, 30, \\[0.4em] 31, 31, 30, 31, 30, 31. \end{gathered}

It has $12$ terms and no simple formula, and that is fine. The perfect squares $1, 4, 9, 16, \dots$ form the infinite sequence $a_n = n^2$ for $n \ge 1$ . A finite sequence with indices from $3$ to $10$ has $10 - 3 + 1 = 8$ terms, not $7$ .

Factorial

Definition For a positive integer $n$ , the factorial of $n$ is the product of the integers from $1$ to $n$ ,

n! = n(n - 1)(n - 2) \cdots 2 \cdot 1.

By definition $0! = 1$ . Factorials are defined here only for nonnegative integers, and for every $n \ge 1$ they satisfy

n! = n \cdot (n - 1)!.

The number $n!$ counts the ways to arrange $n$ different objects in a row.

In plain words Multiply all the whole numbers from $n$ down to $1$ . The value $0! = 1$ is chosen so that the step $n! = n \cdot (n - 1)!$ also works for $n = 1$ .

Example The first factorials grow quickly.

$n$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$n!$	$1$	$1$	$2$	$6$	$24$	$120$	$720$	$5040$

Five different books can stand on a shelf in $5! = 120$ orders. There are $5$ choices for the first spot, then $4$ for the second, and so on down to $1$ for the last. Already $10! = 3\,628\,800$ .

Simplifying factorial expressions

Definition For integers $n > k \ge 0$ , the factors of $n!$ from $n$ down to $k + 1$ can be split off, so

n! = n(n - 1) \cdots (k + 1) \cdot k!

and

\frac{n!}{k!} = n(n - 1) \cdots (k + 1).

The quotient is the product of the $n - k$ consecutive integers from $k + 1$ to $n$ . When $n = k$ the quotient is simply $1$ . Factorials do not distribute over sums or products. For example $(2 \cdot 3)! = 720$ while $2! \cdot 3! = 12$ , and $(3 + 2)! = 120$ while $3! + 2! = 8$ .

In plain words Write the larger factorial out until the smaller one appears, then cancel it. Most of the product disappears.

Example Cancelling $8!$ gives

\frac{10!}{8!} = \frac{10 \cdot 9 \cdot 8!}{8!} = 90.

The same idea works with letters. For $n \ge 0$ and for $n \ge 1$ respectively,

\begin{gathered} \frac{(n + 2)!}{n!} = (n + 2)(n + 1), \\[0.8em] \frac{(2n)!}{(2n - 1)!} = 2n. \end{gathered}

Arithmetic on terms

Definition Let $\{a_n\}$ and $\{b_n\}$ be sequences with the same indices and let $c$ be a real number. Their sum, difference, constant multiple and product are formed term by term,

\begin{gathered} \{a_n + b_n\}, \qquad \{a_n - b_n\}, \\[0.4em] \{c\,a_n\}, \qquad \{a_n b_n\}, \end{gathered}

and the quotient $\{a_n / b_n\}$ is defined when $b_n \ne 0$ for every $n$ . To get a term at a shifted index such as $a_{n+1}$ or $a_{2n}$ , replace every $n$ in the explicit formula by $n + 1$ or by $2n$ . The difference $a_{n+1} - a_n$ and, for nonzero terms, the ratio $a_{n+1} / a_n$ compare each term with the next one.

In plain words Sequences combine position by position, first with first and second with second. A shifted index means substituting into the formula, and the term $a_{n+1}$ is usually not $a_n + 1$ .

Example For $a_n = n^2$ with $n \ge 1$ ,

\begin{gathered} a_{n+1} = (n + 1)^2 = n^2 + 2n + 1, \\[0.4em] a_{n+1} - a_n = 2n + 1, \end{gathered}

so consecutive squares differ by the odd numbers $3, 5, 7, \dots$ , while $a_n + 1 = n^2 + 1$ is something else. For $b_n = 3^n / n!$ with $n \ge 0$ the ratio of consecutive terms is

\frac{b_{n+1}}{b_n} = \frac{3^{n+1}}{(n + 1)!} \cdot \frac{n!}{3^n} = \frac{3}{n + 1}.

This ratio is less than $1$ for $n \ge 3$ , so every term after $b_3$ is smaller than the one before it. The terms $1, 3, 4.5, 4.5, 3.375, 2.025$ show the turn.

Check yourself

Write the first four terms of the sequence below, with $n$ starting at $0$ .

a_n = \frac{(-1)^n}{n!}

Show the answer

The terms are $a_0 = 1$ , $a_1 = -1$ , $a_2 = \frac{1}{2}$ and $a_3 = -\frac{1}{6}$ . Remember that $0! = 1$ and that $(-1)^n$ flips the sign at each step.

Simplify the expressions below, the second one for $n \ge 1$ .

\frac{12!}{10! \cdot 2!}, \qquad \frac{(n + 1)!}{(n - 1)!}

Show the answer

Cancel the smaller factorial in each,

\begin{gathered} \frac{12 \cdot 11 \cdot 10!}{10! \cdot 2} = \frac{132}{2} = 66, \\[0.8em] \frac{(n + 1) \cdot n \cdot (n - 1)!}{(n - 1)!} = n(n + 1). \end{gathered}

For $a_n = \frac{1}{n}$ with $n \ge 1$ , find and simplify $a_{n+1} - a_n$ . What does its sign tell you?

Show the answer

The difference is

\begin{aligned} \frac{1}{n + 1} - \frac{1}{n} &= \frac{n - (n + 1)}{n(n + 1)} \\[0.8em] &= -\frac{1}{n(n + 1)}, \end{aligned}

which is negative for every $n \ge 1$ . So each term is smaller than the one before it.

Introduction to Sequences and Factorials

Key terms

Sequence as a function of an integer index

Indexing convention

Explicit term

Finite and infinite sequences

Factorial

Simplifying factorial expressions

Arithmetic on terms

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus