Sequences and Notation
Textbook Reference
| Primary source | Stewart, Calculus, 9th edition, Section 11.1: “Sequences,” pages 762 to 764 (the definition, notation, and ways of describing a sequence) |
| Open companion | OpenStax Calculus Volume 2, Section 5.1: “Sequences” |
| Companion link | https://openstax.org/books/calculus-volume-2/pages/5-1-sequences |
The OpenStax volume is free and openly licensed. The page numbers above are from Stewart 9th edition, the text used in this course. A sequence has two parts: what one is and how it is written. The limit of a sequence and the special properties of monotonic and bounded sequences come next.
Key idea
You already know what a function is. A sequence is a function in disguise.
A function such as $f(x) = 1/x$ accepts any real number in its domain. A sequence is the same idea with the domain narrowed to the counting numbers $1, 2, 3, \dots$ only. Instead of feeding in $x = 2.7$, you feed in $n = 1$, then $n = 2$, then $n = 3$, and you read off the outputs in order. That ordered list of outputs is the sequence.
The picture to hold in your head is a row of mailboxes. Box number $1$ holds the first term, box number $2$ holds the second term, and so on forever. The label on the box is the position $n$. The contents of the box is the value $a_n$. Nothing about a sequence is harder than a function you have already met; the only new thing is that the inputs are the whole numbers and the outputs come in a fixed order.
Because a sequence is a function on the integers, every tool you learned for functions, including limits, carries over with almost no change. That carryover is what makes the limit of a sequence reachable next.
Prerequisite Check
Sequences sit at the base of the chapter, so the only background needed is from earlier calculus. Make sure each item is solid.
If all five are comfortable, you are ready.
Quick Reference
Definition. An infinite sequence is a list of numbers written in a definite order, \[ a_1, \; a_2, \; a_3, \; \dots, \; a_n, \; \dots \] Formally, a sequence is a function whose domain is the set of positive integers. The output at position $n$ is written $a_n$ (read “a sub n”), called the nth term or the general term.
Notation. The whole sequence is written $\{a_n\}$ or $\{a_n\}_{n=1}^{\infty}$. Unless stated otherwise, $n$ starts at $1$.
Three ways to describe a sequence.
- Explicit formula: a rule for $a_n$ directly in terms of $n$, such as $a_n = \dfrac{1}{2^n}$.
- Recursive formula: a starting term plus a rule that builds each term from earlier ones, such as $a_1 = 1,\; a_2 = 1,\; a_n = a_{n-1} + a_{n-2}$ (the Fibonacci sequence).
- List of terms: write out the first several terms and trust the reader to see the pattern, such as $\tfrac{1}{2}, \tfrac{1}{4}, \tfrac{1}{8}, \dots$
To find a formula from a list. Read three features separately: the numerators, the denominators, and the signs. Build each as a small formula in $n$, then combine. Alternating signs come from a factor of $(-1)^n$ or $(-1)^{n-1}$.
Key Concepts
1. A Sequence Is a Function on the Counting Numbers
Stewart’s definition is short: a sequence is a function $f$ whose domain is the set of positive integers. The only reason it does not look like the functions you are used to is the notation. Where a general function writes $f(n)$, a sequence writes $a_n$. The subscript is doing the exact job the parentheses do.
Because the domain is the integers $1, 2, 3, \dots$ and nothing in between, the graph of a sequence is not a smooth curve. It is a scatter of separated dots, one dot above each whole number on the horizontal axis. The dot above $n = 3$ sits at height $a_3$.
| Position $n$ | $1$ | $2$ | $3$ | $4$ | $5$ |
|---|---|---|---|---|---|
| Term $a_n = \dfrac{n}{n+1}$ | $\dfrac{1}{2}$ | $\dfrac{2}{3}$ | $\dfrac{3}{4}$ | $\dfrac{4}{5}$ | $\dfrac{5}{6}$ |
Plotting those five points shows dots climbing toward the line $y = 1$ but never reaching it. Dots crowding toward a height they never reach is exactly what a limit makes precise.
2. Explicit Formulas
An explicit formula gives $a_n$ directly. You can jump to any term without computing the ones before it.
Example 1 (explicit formula, list the terms). List the first four terms of $\left\{\dfrac{1}{2^n}\right\}$.
Goal. Substitute $n = 1, 2, 3, 4$ into the formula and read off the outputs.
Work. \[ a_1 = \frac{1}{2^1} = \frac{1}{2}, \quad a_2 = \frac{1}{2^2} = \frac{1}{4}, \quad a_3 = \frac{1}{2^3} = \frac{1}{8}, \quad a_4 = \frac{1}{2^4} = \frac{1}{16} \]
Answer: $\dfrac{1}{2}, \; \dfrac{1}{4}, \; \dfrac{1}{8}, \; \dfrac{1}{16}$
Recap. The position $n$ went into the exponent, and the output shrank as $n$ grew. This is the sequence of distances in Zeno’s paradox: walk half the remaining distance to the wall at each step.
(Common error: writing $\dfrac{1}{2}n$ instead of $\dfrac{1}{2^n}$. The $n$ is an exponent here, not a multiplier. Read the formula carefully before substituting.)
Example 2 (a different starting index). List the first four terms of $\left\{\dfrac{n}{n+1}\right\}_{n=2}^{\infty}$.
Goal. The subscript on the brace says start at $n = 2$, not $n = 1$. Substitute $n = 2, 3, 4, 5$.
Work. \[ a_2 = \frac{2}{3}, \quad a_3 = \frac{3}{4}, \quad a_4 = \frac{4}{5}, \quad a_5 = \frac{5}{6} \]
Answer: $\dfrac{2}{3}, \; \dfrac{3}{4}, \; \dfrac{4}{5}, \; \dfrac{5}{6}$
Recap. Always read the starting index. The list “first four terms” still means four numbers, but here they begin at the $n = 2$ slot because that is where the sequence is defined to start.
(Common error: ignoring the lower limit and starting at $n = 1$ out of habit. The notation $\{\,\}_{n=2}^{\infty}$ is an instruction; follow it.)
3. Recursive Formulas
A recursive formula gives one or more starting terms and a rule that builds each new term from the ones before it. You cannot jump ahead; you climb one rung at a time.
Example 3 (Fibonacci sequence). List the first seven terms of the sequence defined by \[ f_1 = 1, \quad f_2 = 1, \quad f_n = f_{n-1} + f_{n-2} \quad \text{for } n \geq 3. \]
Goal. Use the two starting terms, then add the two most recent terms to get each next one.
Work. \[ f_3 = f_2 + f_1 = 1 + 1 = 2 \] \[ f_4 = f_3 + f_2 = 2 + 1 = 3, \quad f_5 = 3 + 2 = 5, \quad f_6 = 5 + 3 = 8, \quad f_7 = 8 + 5 = 13 \]
Answer: $1, \; 1, \; 2, \; 3, \; 5, \; 8, \; 13$
Recap. Each term is the sum of the two before it. This sequence, from a 13th-century rabbit-population problem, is the classic example of a rule that is easy to state recursively but harder to write as a single explicit formula.
(Common error: forgetting that a recursive rule needs enough starting terms. A rule using $f_{n-1}$ and $f_{n-2}$ needs two seeds, here $f_1$ and $f_2$. With only one seed the rule cannot get started.)
4. Finding a Formula From a Pattern
The reverse skill is common on exams: you are handed a list and asked for the general term. Stewart’s worked example is the model. Read three features one at a time.
Example 4 (build the general term). Find a formula for the general term $a_n$ of \[ \frac{3}{5}, \; -\frac{4}{25}, \; \frac{5}{125}, \; -\frac{6}{625}, \; \frac{7}{3125}, \; \dots \] assuming the pattern continues.
Goal. Treat numerator, denominator, and sign as three separate small problems.
Work.
- Numerators are $3, 4, 5, 6, 7, \dots$ starting at $3$ and increasing by $1$. The $n$th numerator is $n + 2$.
- Denominators are $5, 25, 125, 625, 3125, \dots$, which are $5^1, 5^2, 5^3, \dots$. The $n$th denominator is $5^n$.
- Signs alternate, starting positive. A positive first term means $(-1)^{n-1}$ (which is $+1$ when $n = 1$).
Combine the three pieces: \[ a_n = (-1)^{n-1}\,\frac{n+2}{5^n} \]
Answer: $a_n = (-1)^{n-1}\dfrac{n+2}{5^n}$
Recap. Splitting the pattern into numerator, denominator, and sign turns one confusing list into three easy patterns. Always check your formula on $n = 1$ and $n = 2$ before trusting it.
(Common error: using $(-1)^{n}$ when the first term is positive. That factor gives $-1$ at $n = 1$, flipping every sign. When $a_1$ is positive, use $(-1)^{n-1}$ or equivalently $(-1)^{n+1}$.)
Inline Self-Check
Before moving on, try this one in your head.
Question. The sequence $\{\sqrt{n+2}\}_{n=1}^{\infty}$ starts at $n = 1$. What are its first three terms?
Show answer
Substitute $n = 1, 2, 3$: $\sqrt{1+2} = \sqrt{3}$, then $\sqrt{2+2} = \sqrt{4} = 2$, then $\sqrt{3+2} = \sqrt{5}$.
So the first three terms are $\sqrt{3}, \; 2, \; \sqrt{5}$.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Exponent read as a multiplier | $\dfrac{1}{2^n}$ treated as $\dfrac{1}{2}n$ | The $n$ is in the exponent; compute $2^n$ first |
| Ignoring the starting index | Starting $\{a_n\}_{n=2}^{\infty}$ at $n = 1$ | Begin substitution at the index on the brace |
| Wrong sign factor | Using $(-1)^n$ when $a_1 > 0$ | Use $(-1)^{n-1}$ so the first term is positive |
| Too few seeds for a recursion | One seed for a rule using two prior terms | Supply as many starting terms as the rule references |
| Confusing position with value | Reporting $n$ instead of $a_n$ | The answer is the term $a_n$, not the slot number |
Common Misconceptions
a sequence is a function of all real numbers, just with integer inputs labeled differently.
This is the action-view-of-function error. A sequence is defined only on the positive integers; there is no value between $a_3$ and $a_4$. The graph consists of isolated dots, not a continuous curve. Treating the formula as a continuous function and evaluating it at, say, $n = 2.5$ is outside the domain and has no meaning in the definition of the sequence.
the alternating factor $(-1)^n$ makes the first term negative when $a_1$ should be positive.
This is the input-output-confusion error in sign factors. The factor $(-1)^n$ evaluates to $-1$ when $n = 1$, making the first term negative. When the listed sequence starts with a positive term, the correct factor is $(-1)^{n-1}$ or equivalently $(-1)^{n+1}$, each of which equals $+1$ at $n = 1$.
Leveled Practice
Attempt each problem before opening the answer.
Level 1: Direct Application
Problem 1. List the first five terms of $a_n = \dfrac{n}{2n+1}$.
Show answer
\[ a_1 = \frac{1}{3}, \quad a_2 = \frac{2}{5}, \quad a_3 = \frac{3}{7}, \quad a_4 = \frac{4}{9}, \quad a_5 = \frac{5}{11} \]
Problem 2. List the first four terms of $a_n = \dfrac{(-1)^n}{n^2}$.
Show answer
\[ a_1 = \frac{-1}{1} = -1, \quad a_2 = \frac{1}{4}, \quad a_3 = -\frac{1}{9}, \quad a_4 = \frac{1}{16} \]
The $(-1)^n$ factor makes the odd terms negative and the even terms positive.
Problem 3. A recursive sequence is defined by $a_1 = 1$ and $a_{n+1} = 2a_n + 1$. List the first four terms.
Show answer
\[ a_1 = 1, \quad a_2 = 2(1) + 1 = 3, \quad a_3 = 2(3) + 1 = 7, \quad a_4 = 2(7) + 1 = 15 \]
Level 2: Finding the Formula
Problem 4. Find an explicit formula for the general term of $\dfrac{1}{1}, \dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{4}, \dots$ (starting at $n = 1$).
Show answer
The numerators are all $1$ and the denominators are $1, 2, 3, 4, \dots = n$. So \[ a_n = \frac{1}{n}. \]
Problem 5. Find an explicit formula for $-\dfrac{1}{2}, \dfrac{2}{4}, -\dfrac{3}{8}, \dfrac{4}{16}, \dots$ (starting at $n = 1$).
Show answer
Numerators: $1, 2, 3, 4, \dots = n$. Denominators: $2, 4, 8, 16, \dots = 2^n$. Signs start negative, so use $(-1)^n$. Therefore \[ a_n = (-1)^n\,\frac{n}{2^n}. \]
Check $n = 1$: $(-1)^1 \cdot \tfrac{1}{2} = -\tfrac{1}{2}$. Correct.
Level 3: Deeper Problems
Problem 6. The terms of a sequence are $\sqrt{2}, \; \sqrt{2\sqrt{2}}, \; \sqrt{2\sqrt{2\sqrt{2}}}, \dots$. Write a recursive formula that generates this sequence.
Show answer
Each term takes the previous term, multiplies by $2$ inside a square root. Start with $a_1 = \sqrt{2}$ and use \[ a_{n+1} = \sqrt{2\,a_n}. \]
Check: $a_2 = \sqrt{2\sqrt{2}}$ and $a_3 = \sqrt{2\sqrt{2\sqrt{2}}}$, matching the list.
Problem 7. Explain in one or two sentences why the sequence $a_n = (-1)^n$ has no single explicit “value it heads toward,” using only the idea of the term list.
Show answer
Writing the terms out gives $-1, 1, -1, 1, \dots$. The list never settles on one number; it jumps back and forth between $-1$ and $1$ forever. A sequence heads toward a value only when its terms close in on that one value, and these do not.
(The limit of a sequence makes this idea of “heading toward a value” precise.)
Mastery Checklist
You have mastered this skill when you can do all of the following without notes:
Mental Model
Picture an infinite row of numbered mailboxes stretching to the right. The label on each box is its position $n$. Inside each box is one number, the term $a_n$.
An explicit formula is a master key: hand it any box number and it tells you the contents of that box directly, without opening any other box. A recursive formula is a chain of keys: each box can be opened only after the one or two boxes before it, because its contents are built from theirs.
The graph of the sequence is just the heights of the contents, one dot above each box. Because there is a box only at each whole number, the dots are separated, never a connected curve. The boxes-and-dots picture makes the limit almost obvious: a sequence “approaches” a value when the dots crowd toward one height as you walk down the row of boxes.
Connections
Built From
- Function notation and evaluation. A sequence is a function on the integers; $a_n$ plays the role of $f(n)$. Every skill of substituting a value into a formula is reused here.
Leads To
- The limit of a sequence: the precise meaning of the terms “approaching” a value, built directly on the dot-crowding picture.
- Properties of sequences: monotonic (always increasing or always decreasing) and bounded sequences, and the theorem that ties those two properties to convergence.
- Infinite series: adding the terms of a sequence. A series is built on top of a sequence, so this notation underlies every later topic in the chapter.
Why It Matters
Sequences are how mathematics records a process that happens in stages: successive approximations from Newton’s method, the balance in an account after each year of compound interest, the population of a species in each new generation. Any time a quantity is updated step by step, a sequence is the natural language for it. In computing, this is exactly an indexed array or the successive states of a loop, so the idea transfers directly to programming.
Audience Notes
For students who find math intimidating: there is nothing new to fear here. If you can plug a number into a formula, you can list the terms of a sequence. The notation $a_n$ is just a new coat of paint on $f(n)$.
For students who want depth: the formal definition (a function with domain the positive integers) is worth taking seriously, because it is what lets every theorem about limits of functions transfer to limits of sequences with only a change of domain. That transfer is what drives the limit of a sequence.
For students aimed at a career in computing or engineering: a recursive formula is a recurrence relation, the same object that governs the running time of recursive algorithms and the behavior of discrete-time systems. The Fibonacci recurrence here is the textbook first example in both fields.
Back to Sequences, Series, and Power Series | Next: Limit of a Sequence