Sequences and Their Properties: Learn It 1

  • Find the pattern in a sequence and write a formula for its terms
  • Determine what value a sequence approaches (if it approaches any value at all)
  • Figure out if a sequence converges or diverges

What is a Sequence?

A sequence is simply an ordered list of numbers that follows a pattern. You encounter sequences regularly—think of your monthly rent payments, the distances you run each week, or even the Fibonacci numbers. Understanding sequences helps us analyze patterns and predict future behavior in mathematical models.

infinite sequence

An infinite sequence [latex]{a_n}[/latex] is an ordered list of numbers:

[latex]a_1, a_2, a_3, \ldots, a_n, \ldots[/latex]

Key terms:

  • Each number [latex]a_n[/latex] is called a term
  • The subscript [latex]n[/latex] is the index variable
  • We write sequences as [latex]{a_n}_{n=1}^{\infty}[/latex] or simply [latex]{a_n}[/latex]
Important: A similar notation is used for sets, but a sequence is an ordered list (like a line at the grocery store), while a set is unordered collection (like items in a shopping basket). Order matters for sequences!

Consider this sequence: [latex]2, 4, 8, 16, 32, \ldots[/latex]

You can see that [latex]a_1 = 2[/latex], [latex]a_2 = 4[/latex], [latex]a_3 = 8[/latex], and so on. Each term is a power of 2. But how do we capture this pattern mathematically? There are two main approaches.

Method 1: Explicit Formula

An explicit formula gives you [latex]a_n[/latex] directly in terms of [latex]n[/latex].

For our sequence: [latex]a_n = 2^n[/latex]

This means [latex]a_1 = 2^1 = 2[/latex], [latex]a_2 = 2^2 = 4[/latex], [latex]a_3 = 2^3 = 8[/latex], and so on. The beauty of an explicit formula is that you can find any term directly. Want the 100th term? Just calculate [latex]a_{100} = 2^{100}[/latex] without computing all the previous terms.

Method 2: Recurrence Relation

A recurrence relation defines each term using previous terms.

For the same sequence: [latex]a_1 = 2[/latex] and [latex]a_n = 2a_{n-1}[/latex] for [latex]n \geq 2[/latex]

This says “start with 2, then each new term is twice the previous term.”

Recurrence relations are particularly useful when each term naturally depends on what came before—like population growth where next year’s population depends on this year’s. However, finding a specific term requires calculating all the previous ones.

Sequences don’t have to start at [latex]n = 1[/latex]. You might see sequences starting at [latex]n = 0[/latex] or any other integer, depending on the context. Just pay attention to the index range given.

Since a sequence [latex]{a_n}[/latex] assigns exactly one value to each positive integer [latex]n[/latex], we can think of it as a function whose domain is the set of positive integers. This means we can graph sequences just like we graph functions.

The graph of sequence [latex]{a_n}[/latex] consists of discrete points [latex](n, a_n)[/latex] for all positive integers [latex]n[/latex]. Unlike continuous functions, sequence graphs show isolated dots rather than connected curves.

A graph in quadrant one containing the following points: (1, 2), (2, 4), (3, 8), (4, 16).
Figure 1. The plotted points are a graph of the sequence [latex]\left\{{2}^{n}\right\}[/latex].

Special Types of Sequences

Two types of sequences appear frequently in mathematics and have special names because of their predictable patterns.

Arithmetic Sequences

In an arithmetic sequence, the difference between every pair of consecutive terms is the same.

arithmetic sequence

An arithmetic sequence has the same difference between consecutive terms.

General form: [latex]a_n = cn + b[/latex] (explicit formula)

The common difference is [latex]d = a_{n+1} - a_n[/latex]

Consider the sequence [latex]3, 7, 11, 15, 19, \ldots[/latex]

Notice that each term increases by 4: [latex]7-3=4[/latex], [latex]11-7=4[/latex], [latex]15-11=4[/latex], and so on.

We can describe this sequence using a recurrence relation: [latex]a_1 = 3[/latex] and [latex]a_n = a_{n-1} + 4[/latex] for [latex]n \geq 2[/latex]

Or we can find the explicit formula by observing the pattern:

  • [latex]a_2 = 3 + 4[/latex]
  • [latex]a_3 = 3 + 4 + 4 = 3 + 2 \cdot 4[/latex]
  • [latex]a_4 = 3 + 4 + 4 + 4 = 3 + 3 \cdot 4[/latex]

This gives us the explicit formula: [latex]a_n = 3 + 4(n-1) = 4n - 1[/latex]

Geometric Sequences

In a geometric sequence, the ratio of every pair of consecutive terms is the same.

geometric sequence

A geometric sequence has the same ratio between consecutive terms.

General form: [latex]a_n = cr^n[/latex] (explicit formula)

The common ratio is [latex]r = \frac{a_{n+1}}{a_n}[/latex]

Consider the sequence [latex]2, -\frac{2}{3}, \frac{2}{9}, -\frac{2}{27}, \frac{2}{81}, \ldots[/latex]

Each term is [latex]-\frac{1}{3}[/latex] times the previous term. We can verify: [latex]\frac{-\frac{2}{3}}{2} = -\frac{1}{3}[/latex] and [latex]\frac{\frac{2}{9}}{-\frac{2}{3}} = -\frac{1}{3}[/latex].

Using a recurrence relation: [latex]a_1 = 2[/latex] and [latex]a_n = -\frac{1}{3} \cdot a_{n-1}[/latex] for [latex]n \geq 2[/latex]

Or we can find the explicit formula by following the pattern:

  • [latex]a_2 = -\frac{1}{3} \cdot 2[/latex]
  • [latex]a_3 = \left(-\frac{1}{3}\right)^2 \cdot 2[/latex]
  • [latex]a_4 = \left(-\frac{1}{3}\right)^3 \cdot 2[/latex]

This gives us: [latex]a_n = 2\left(-\frac{1}{3}\right)^{n-1}[/latex]

Quick Recognition

  • Arithmetic: Look for the same number being added (or subtracted) repeatedly
  • Geometric: Look for the same number being multiplied repeatedly

For each of the following sequences, find an explicit formula for the [latex]n\text{th}[/latex] term of the sequence.

  1. [latex]-\frac{1}{2},\frac{2}{3},-\frac{3}{4},\frac{4}{5},-\frac{5}{6}\text{,}\ldots[/latex]
  2. [latex]\frac{3}{4},\frac{9}{7},\frac{27}{10},\frac{81}{13},\frac{243}{16}\text{,}\ldots[/latex]

For each of the following recursively defined sequences, find an explicit formula for the sequence.

  1. [latex]{a}_{1}=2[/latex], [latex]{a}_{n}=-3{a}_{n - 1}[/latex] for [latex]n\ge 2[/latex]
  2. [latex]{a}_{1}=\frac{1}{2}[/latex], [latex]{a}_{n}={a}_{n - 1}+{\left(\frac{1}{2}\right)}^{n}[/latex] for [latex]n\ge 2[/latex]