Chapter 8: Sequences & Series
Sequences and series are among the most elegant ideas in algebra. A sequence is simply an ordered list of numbers that follows a pattern, and a series is what you get when you add those numbers together. These concepts appear everywhere: in calculating loan payments, modeling population growth, analyzing investment returns, and even in the mathematics behind music and architecture. In this chapter you will develop the tools to identify, describe, and sum both arithmetic and geometric patterns, and you will see how infinite sums can converge to finite values.
Table of Contents
8.1 Sequences & Notation
We begin with the fundamental language of sequences. In everyday life you encounter sequences constantly: house numbers along a street (2, 4, 6, 8, ...), the number of petals on successive rings of a sunflower, or the balance of a savings account month after month. Mathematically, a sequence is a function whose domain is the set of positive integers (or sometimes the non-negative integers).
A sequence is an ordered list of numbers $a_1, a_2, a_3, \ldots$ where each term $a_n$ is associated with a positive integer $n$ called its index. We write $\{a_n\}$ or $\{a_n\}_{n=1}^{\infty}$ to denote the entire sequence.
For example, the sequence $3, 7, 11, 15, 19, \ldots$ has first term $a_1 = 3$, second term $a_2 = 7$, and so on. There are two primary ways to describe how a sequence is generated.
Explicit Formulas
An explicit formula (also called a closed-form formula) gives $a_n$ directly as a function of $n$. You can compute any term without knowing the previous terms.
The sequence defined by $a_n = 4n - 1$ produces:
$a_1 = 4(1) - 1 = 3, \quad a_2 = 4(2) - 1 = 7, \quad a_3 = 4(3) - 1 = 11, \quad a_{10} = 4(10) - 1 = 39$
Notice that we can jump straight to $a_{10}$ without computing the nine terms before it.
Recursive Formulas
A recursive formula defines each term in relation to one or more previous terms, together with an initial condition. To find $a_{10}$ you must first compute $a_1$ through $a_9$.
Let $a_1 = 3$ and $a_n = a_{n-1} + 4$ for $n \geq 2$. Then:
$a_1 = 3, \quad a_2 = 3 + 4 = 7, \quad a_3 = 7 + 4 = 11, \quad a_4 = 11 + 4 = 15$
This recursive definition produces the same sequence as the explicit formula $a_n = 4n - 1$ from Example 1.
Sigma Notation
When we want to add up terms of a sequence, we use sigma notation (also called summation notation). The Greek capital letter sigma $\Sigma$ tells us to sum:
$$\sum_{k=1}^{n} a_k = a_1 + a_2 + a_3 + \cdots + a_n$$Here $k$ is the index of summation, $1$ is the lower limit, and $n$ is the upper limit. The variable $k$ is a dummy variable -- you can replace it with any letter without changing the value of the sum.
Evaluate $\displaystyle\sum_{k=1}^{5} (2k + 1)$.
Solution: Substitute $k = 1, 2, 3, 4, 5$:
$(2(1)+1) + (2(2)+1) + (2(3)+1) + (2(4)+1) + (2(5)+1) = 3 + 5 + 7 + 9 + 11 = 35$
Key properties of sigma notation that simplify calculations:
- Constant multiple: $\displaystyle\sum_{k=1}^{n} c \cdot a_k = c \sum_{k=1}^{n} a_k$
- Sum/difference: $\displaystyle\sum_{k=1}^{n}(a_k \pm b_k) = \sum_{k=1}^{n} a_k \pm \sum_{k=1}^{n} b_k$
- Constant sum: $\displaystyle\sum_{k=1}^{n} c = cn$
8.2 Arithmetic Sequences
An arithmetic sequence is the simplest type of pattern: each term is obtained by adding the same fixed number to the previous term. The temperature dropping by 3 degrees every hour, a salary increasing by $2{,}000$ each year, or the seats in successive rows of an auditorium increasing by a constant amount are all arithmetic sequences.
A sequence $\{a_n\}$ is arithmetic if there exists a constant $d$ (called the common difference) such that $a_{n} - a_{n-1} = d$ for all $n \geq 2$. Equivalently, $a_n = a_{n-1} + d$.
The $n$th term of an arithmetic sequence with first term $a_1$ and common difference $d$ is:
$$a_n = a_1 + (n - 1)d$$
This formula is linear in $n$: if you plot the terms of an arithmetic sequence, the points lie on a straight line with slope $d$ and $y$-intercept $a_1 - d$.
An arithmetic sequence begins $7, 13, 19, 25, \ldots$ Find $a_{50}$.
Solution: The common difference is $d = 13 - 7 = 6$. Using the explicit formula:
$$a_{50} = 7 + (50 - 1)(6) = 7 + 294 = 301$$
Finding $d$ and $a_1$ from Two Terms
If you know any two terms $a_m$ and $a_p$ of an arithmetic sequence (where $m \neq p$), you can recover both the common difference and the first term.
$$d = \frac{a_m - a_p}{m - p}$$
Once $d$ is known, solve for $a_1$ using $a_1 = a_m - (m-1)d$.
An arithmetic sequence has $a_4 = 23$ and $a_{10} = 53$. Find $a_1$ and $d$, then write the explicit formula.
Solution:
$$d = \frac{a_{10} - a_4}{10 - 4} = \frac{53 - 23}{6} = 5$$
$$a_1 = a_4 - 3d = 23 - 15 = 8$$
The explicit formula is $a_n = 8 + (n-1)(5) = 5n + 3$.
8.3 Arithmetic Series
A series is the sum of the terms of a sequence. An arithmetic series is the sum of the terms of an arithmetic sequence. The famous story of young Gauss summing the integers from 1 to 100 illustrates the key insight: you can pair terms from opposite ends to compute the sum efficiently.
The arithmetic series $S_n$ is the sum of the first $n$ terms of an arithmetic sequence: $S_n = a_1 + a_2 + \cdots + a_n$.
Gauss's trick works because $a_1 + a_n = a_2 + a_{n-1} = a_3 + a_{n-2} = \cdots$ in any arithmetic sequence. There are $n$ terms total, forming $n/2$ such pairs, each summing to $a_1 + a_n$.
$$S_n = \frac{n}{2}(a_1 + a_n) = \frac{n}{2}\bigl(2a_1 + (n-1)d\bigr)$$
The two forms are equivalent; the first is convenient when you know the last term $a_n$, and the second when you know $d$ but not $a_n$.
Find $1 + 2 + 3 + \cdots + 100$.
Solution: Here $a_1 = 1$, $a_{100} = 100$, and $n = 100$:
$$S_{100} = \frac{100}{2}(1 + 100) = 50 \times 101 = 5{,}050$$
A stadium section has 20 rows. The first row has 15 seats and each subsequent row has 3 more seats than the previous one. How many seats are in the section?
Solution: This is an arithmetic series with $a_1 = 15$, $d = 3$, and $n = 20$.
$$S_{20} = \frac{20}{2}\bigl(2(15) + (20-1)(3)\bigr) = 10(30 + 57) = 10 \times 87 = 870 \text{ seats}$$
Writing a Series in Sigma Notation
The arithmetic series from Example 7 can be written compactly as:
$$\sum_{k=1}^{20}(15 + (k-1) \cdot 3) = \sum_{k=1}^{20}(3k + 12)$$This notation makes the structure of the sum explicit and is essential for later work in calculus.
8.4 Geometric Sequences
While arithmetic sequences grow by repeated addition, geometric sequences grow by repeated multiplication. This makes them the natural model for exponential growth and decay: population doubling, radioactive half-life, compound interest, and the bounce heights of a ball are all geometric sequences.
A sequence $\{a_n\}$ is geometric if there exists a nonzero constant $r$ (called the common ratio) such that $\frac{a_n}{a_{n-1}} = r$ for all $n \geq 2$. Equivalently, $a_n = a_{n-1} \cdot r$.
The $n$th term of a geometric sequence with first term $a_1$ and common ratio $r$ is:
$$a_n = a_1 \cdot r^{n-1}$$
Unlike the linear graph of arithmetic sequences, the graph of a geometric sequence follows an exponential curve. When $|r| > 1$ the terms grow without bound; when $0 < |r| < 1$ the terms shrink toward zero; when $r < 0$ the terms alternate in sign.
A culture of bacteria starts with 500 cells and triples every hour. Write an explicit formula and find the population after 6 hours.
Solution: Here $a_1 = 500$ and $r = 3$. The explicit formula is $a_n = 500 \cdot 3^{n-1}$.
$$a_7 = 500 \cdot 3^6 = 500 \cdot 729 = 364{,}500 \text{ cells}$$
(Note: $a_7$ represents the population at the end of the 6th hour, since $a_1$ is time $t=0$.)
A ball is dropped from 80 feet and bounces to $\frac{3}{4}$ of its previous height each time. Find the height of the 5th bounce.
Solution: The bounce heights form a geometric sequence: $a_1 = 80 \cdot \frac{3}{4} = 60$, with $r = \frac{3}{4}$.
$$a_5 = 60 \cdot \left(\frac{3}{4}\right)^{4} = 60 \cdot \frac{81}{256} = \frac{4860}{256} \approx 18.98 \text{ feet}$$
Finding the Common Ratio
To find $r$, divide any term by the preceding term: $r = \frac{a_{n}}{a_{n-1}}$. If you are given two non-consecutive terms $a_m$ and $a_p$, you can find $r$ by solving:
$$a_m = a_p \cdot r^{m-p} \implies r = \left(\frac{a_m}{a_p}\right)^{1/(m-p)}$$A geometric sequence has $a_2 = 12$ and $a_5 = 324$. Find $r$ and $a_1$.
Solution:
$$r^{5-2} = \frac{a_5}{a_2} = \frac{324}{12} = 27 \implies r^3 = 27 \implies r = 3$$
$$a_1 = \frac{a_2}{r} = \frac{12}{3} = 4$$
Interactive: Arithmetic vs. Geometric Sequences
Use the graph below to compare arithmetic and geometric sequences side by side. Adjust $a_1$ (the first term), $d$ (the common difference for the arithmetic sequence), and $r$ (the common ratio for the geometric sequence) using the sliders. Notice how the arithmetic sequence forms a straight line while the geometric sequence curves exponentially.
Interactive: Drag sliders to adjust $a_1$, $d$, and $r$. Blue dots = arithmetic sequence; Red dots = geometric sequence.
8.5 Geometric Series
Adding the terms of a geometric sequence produces a geometric series. The sum formula is derived by a clever algebraic trick: multiply the series by $r$ and subtract to collapse most terms.
Finite Geometric Series
For $r \neq 1$, the sum of the first $n$ terms of a geometric sequence is:
$$S_n = a_1 \cdot \frac{1 - r^n}{1 - r}$$
Derivation sketch: Let $S_n = a_1 + a_1 r + a_1 r^2 + \cdots + a_1 r^{n-1}$. Multiply both sides by $r$: $rS_n = a_1 r + a_1 r^2 + \cdots + a_1 r^n$. Subtract: $S_n - rS_n = a_1 - a_1 r^n$, so $S_n(1 - r) = a_1(1 - r^n)$, giving the formula above.
Find $\displaystyle\sum_{k=1}^{8} 5 \cdot 2^{k-1}$.
Solution: Here $a_1 = 5$, $r = 2$, and $n = 8$:
$$S_8 = 5 \cdot \frac{1 - 2^8}{1 - 2} = 5 \cdot \frac{1 - 256}{-1} = 5 \cdot 255 = 1{,}275$$
Infinite Geometric Series
When $|r| < 1$, each successive term becomes smaller and smaller. The partial sums $S_1, S_2, S_3, \ldots$ approach a finite limit as $n \to \infty$, because $r^n \to 0$. This allows us to define the sum of an infinite geometric series.
If $|r| < 1$, then the infinite geometric series converges and its sum is:
$$S = \frac{a_1}{1 - r}$$
If $|r| \geq 1$, the series diverges (does not have a finite sum).
This result is remarkable: an infinite number of positive terms can add up to a finite number, provided the terms shrink fast enough.
Evaluate $\displaystyle\sum_{k=0}^{\infty} 12 \cdot \left(\frac{1}{3}\right)^k$.
Solution: Here $a_1 = 12$ (when $k=0$) and $r = \frac{1}{3}$. Since $|r| = \frac{1}{3} < 1$:
$$S = \frac{12}{1 - \frac{1}{3}} = \frac{12}{\frac{2}{3}} = 18$$
Express $0.272727\ldots$ as a fraction using geometric series.
Solution: Write $0.\overline{27} = \frac{27}{100} + \frac{27}{10{,}000} + \frac{27}{1{,}000{,}000} + \cdots$
This is a geometric series with $a_1 = \frac{27}{100}$ and $r = \frac{1}{100}$:
$$S = \frac{27/100}{1 - 1/100} = \frac{27/100}{99/100} = \frac{27}{99} = \frac{3}{11}$$
Interactive: Partial Sum Convergence
The graph below plots the partial sums $S_n = a_1 \cdot \frac{1 - r^n}{1 - r}$ of a geometric series. When $|r| < 1$, watch how the partial sums level off toward the limit $\frac{a_1}{1-r}$ (shown as a dashed line). Try changing $r$ to values close to 1 or to negative values to see how the convergence behavior changes.
Interactive: Adjust $a_1$ and $r$ (with $|r| < 1$) to see partial sums converge to $\frac{a_1}{1-r}$. The dashed green line shows the infinite sum.
8.6 Applications
Sequences and series are not just abstract algebra -- they model real situations across finance, biology, physics, and computer science. Here we explore several important applications.
Compound Interest as a Geometric Sequence
When money earns compound interest, the balance at the end of each compounding period forms a geometric sequence. If you invest a principal $P$ at an annual interest rate $r$ compounded once per year, the balance after $n$ years is:
$$A_n = P(1 + r)^n$$This is a geometric sequence with first term $a_0 = P$ and common ratio $(1 + r)$.
You invest $\$5{,}000$ in an account that earns 6% annual interest, compounded annually. What is the balance after 10 years?
Solution:
$$A_{10} = 5000(1.06)^{10} = 5000 \times 1.7908\ldots \approx \$8{,}954.24$$
For monthly compounding at annual rate $r$, the monthly rate is $r/12$, and after $n$ months the balance is $A_n = P\left(1 + \frac{r}{12}\right)^n$.
Annuities: Series of Regular Payments
When you make regular deposits (an annuity), the total value is a geometric series. If you deposit $D$ dollars at the end of each period at rate $i$ per period for $n$ periods, the future value is:
$$FV = D \cdot \frac{(1+i)^n - 1}{i}$$You save $\$200$ per month in an account earning 5% annual interest compounded monthly. How much do you have after 30 years?
Solution: Monthly rate $i = 0.05/12 \approx 0.004167$, and $n = 360$ months.
$$FV = 200 \cdot \frac{(1.004167)^{360} - 1}{0.004167} = 200 \cdot \frac{4.4677 - 1}{0.004167} \approx 200 \times 832.26 \approx \$166{,}452$$
Fibonacci-like Recursions
Not all recursive sequences are arithmetic or geometric. The famous Fibonacci sequence is defined by:
$$F_1 = 1, \quad F_2 = 1, \quad F_n = F_{n-1} + F_{n-2} \text{ for } n \geq 3$$This produces $1, 1, 2, 3, 5, 8, 13, 21, 34, 55, \ldots$ The Fibonacci sequence appears in nature (phyllotaxis in plants, spiral shell growth) and has the remarkable property that the ratio of consecutive terms converges to the golden ratio $\phi = \frac{1 + \sqrt{5}}{2} \approx 1.618$.
Define $a_1 = 2$, $a_2 = 5$, and $a_n = a_{n-1} + a_{n-2}$ for $n \geq 3$. Find $a_7$.
Solution: Build the sequence term by term:
$a_3 = 5 + 2 = 7, \quad a_4 = 7 + 5 = 12, \quad a_5 = 12 + 7 = 19$
$a_6 = 19 + 12 = 31, \quad a_7 = 31 + 19 = 50$
Real-World Modeling
Arithmetic sequences model situations with constant change (linear growth or decay), while geometric sequences model situations with constant percentage change (exponential growth or decay).
- Arithmetic: A taxi charges $\$3.00$ base fare plus $\$2.50$ per mile. The cost after $n$ miles is $a_n = 3 + 2.5n$.
- Geometric: A car depreciates by 15% each year. After $n$ years, its value is $V_n = V_0 \cdot (0.85)^n$.
- Arithmetic series: The total distance a freely falling object covers in $n$ seconds (ignoring air resistance) involves summing an arithmetic sequence of velocity increments.
- Geometric series: The total distance a bouncing ball travels (up and down) before coming to rest is an infinite geometric series.
A ball is dropped from 10 meters and bounces to 60% of its previous height each time. Find the total vertical distance traveled before the ball comes to rest.
Solution: The ball falls 10 m, then bounces up $10(0.6) = 6$ m and falls 6 m, then bounces up $10(0.6)^2 = 3.6$ m and falls 3.6 m, and so on.
Total distance $= 10 + 2\displaystyle\sum_{k=1}^{\infty} 10(0.6)^k = 10 + 2 \cdot \frac{10(0.6)}{1 - 0.6} = 10 + 2 \cdot 15 = 40$ meters.
8.7 Practice Problems
Test your understanding with the problems below. Each problem has a detailed solution that you can reveal by clicking "Show Solution."
Write the first five terms of the sequence defined by $a_n = (-1)^n \cdot \frac{n}{n+1}$.
Show Solution
Substitute $n = 1, 2, 3, 4, 5$:
$a_1 = (-1)^1 \cdot \frac{1}{2} = -\frac{1}{2}$
$a_2 = (-1)^2 \cdot \frac{2}{3} = \frac{2}{3}$
$a_3 = (-1)^3 \cdot \frac{3}{4} = -\frac{3}{4}$
$a_4 = (-1)^4 \cdot \frac{4}{5} = \frac{4}{5}$
$a_5 = (-1)^5 \cdot \frac{5}{6} = -\frac{5}{6}$
The first five terms are $-\frac{1}{2},\; \frac{2}{3},\; -\frac{3}{4},\; \frac{4}{5},\; -\frac{5}{6}$.
Determine whether the sequence $5, 11, 17, 23, 29, \ldots$ is arithmetic, geometric, or neither. If arithmetic, find $d$; if geometric, find $r$.
Show Solution
Check for a common difference: $11 - 5 = 6$, $17 - 11 = 6$, $23 - 17 = 6$.
The sequence is arithmetic with common difference $d = 6$.
Check: the ratios $11/5 = 2.2$ and $17/11 \approx 1.545$ are not equal, confirming it is not geometric.
An arithmetic sequence has $a_3 = 14$ and $a_8 = 39$. Find $a_1$, $d$, and $a_{20}$.
Show Solution
$d = \frac{a_8 - a_3}{8 - 3} = \frac{39 - 14}{5} = 5$
$a_1 = a_3 - 2d = 14 - 10 = 4$
$a_{20} = 4 + (20-1)(5) = 4 + 95 = 99$
Find the sum: $\displaystyle\sum_{k=1}^{40}(3k + 2)$.
Show Solution
This is an arithmetic series. When $k=1$: $a_1 = 5$. When $k=40$: $a_{40} = 122$. Common difference $d = 3$.
$$S_{40} = \frac{40}{2}(5 + 122) = 20 \times 127 = 2{,}540$$
Alternatively: $\displaystyle\sum_{k=1}^{40}(3k + 2) = 3\sum_{k=1}^{40}k + \sum_{k=1}^{40}2 = 3 \cdot \frac{40 \cdot 41}{2} + 80 = 2{,}460 + 80 = 2{,}540$.
A geometric sequence has $a_1 = 4$ and $r = -\frac{1}{2}$. Find $a_6$ and $S_6$.
Show Solution
$a_6 = 4 \cdot \left(-\frac{1}{2}\right)^5 = 4 \cdot \left(-\frac{1}{32}\right) = -\frac{1}{8}$
$$S_6 = 4 \cdot \frac{1 - (-\frac{1}{2})^6}{1 - (-\frac{1}{2})} = 4 \cdot \frac{1 - \frac{1}{64}}{\frac{3}{2}} = 4 \cdot \frac{63/64}{3/2} = 4 \cdot \frac{63}{96} = 4 \cdot \frac{21}{32} = \frac{21}{8} = 2.625$$
Find the sum of the infinite geometric series: $27 + 9 + 3 + 1 + \frac{1}{3} + \cdots$
Show Solution
Here $a_1 = 27$ and $r = \frac{9}{27} = \frac{1}{3}$. Since $|r| = \frac{1}{3} < 1$, the series converges:
$$S = \frac{27}{1 - \frac{1}{3}} = \frac{27}{\frac{2}{3}} = \frac{81}{2} = 40.5$$
Express $0.\overline{63}$ (i.e., $0.636363\ldots$) as a fraction in lowest terms.
Show Solution
$0.\overline{63} = \frac{63}{100} + \frac{63}{10{,}000} + \frac{63}{1{,}000{,}000} + \cdots$
This is a geometric series with $a_1 = \frac{63}{100}$ and $r = \frac{1}{100}$:
$$S = \frac{63/100}{1 - 1/100} = \frac{63/100}{99/100} = \frac{63}{99} = \frac{7}{11}$$
You deposit $\$500$ per month into a retirement account earning 4.8% annual interest compounded monthly. How much will you have after 25 years?
Show Solution
Monthly rate: $i = 0.048/12 = 0.004$. Number of deposits: $n = 25 \times 12 = 300$.
$$FV = 500 \cdot \frac{(1.004)^{300} - 1}{0.004}$$
$(1.004)^{300} = e^{300 \ln(1.004)} = e^{300 \times 0.003992} = e^{1.1976} \approx 3.3117$
$$FV = 500 \cdot \frac{3.3117 - 1}{0.004} = 500 \times 577.93 \approx \$288{,}964$$
Write a recursive formula for the sequence $2, 6, 18, 54, 162, \ldots$ and use it to find $a_8$.
Show Solution
The common ratio is $r = 6/2 = 3$. The recursive formula is:
$a_1 = 2, \quad a_n = 3a_{n-1}$ for $n \geq 2$.
Using the explicit formula to find $a_8$ directly:
$a_8 = 2 \cdot 3^{7} = 2 \cdot 2{,}187 = 4{,}374$
A theater has 30 rows. The first row has 20 seats, and each row has 2 more seats than the row in front of it. How many total seats are in the theater?
Show Solution
This is an arithmetic series with $a_1 = 20$, $d = 2$, $n = 30$.
$a_{30} = 20 + (30-1)(2) = 20 + 58 = 78$ seats in the last row.
$$S_{30} = \frac{30}{2}(20 + 78) = 15 \times 98 = 1{,}470 \text{ seats}$$
Determine whether $\displaystyle\sum_{k=1}^{\infty} 5 \cdot \left(\frac{4}{3}\right)^{k-1}$ converges or diverges. If it converges, find its sum.
Show Solution
Here $r = \frac{4}{3}$ and $|r| = \frac{4}{3} > 1$.
Since $|r| \geq 1$, the series diverges. There is no finite sum.
The first three terms of a Fibonacci-like sequence are $a_1 = 1$, $a_2 = 3$, with $a_n = a_{n-1} + a_{n-2}$. Find $a_{10}$ and the sum $a_1 + a_2 + \cdots + a_{10}$.
Show Solution
Build the sequence: $1, 3, 4, 7, 11, 18, 29, 47, 76, 123$
So $a_{10} = 123$.
Sum: $1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = 319$.