Patterns that grow, shrink, or repeat are everywhere: a savings plan that adds the same amount each month, a ball that loses height with every bounce, a loan balance that changes by a fixed percent. In this chapter you will learn to describe such patterns with sequences, to add up their terms with series, and to write long sums compactly with sigma notation.
1. What is a sequence?
A sequence is an ordered list of numbers called terms. We write \(a_1, a_2, a_3, \dots\) where \(a_1\) is the first term and \(a_n\) is the term in position \(n\) (the general term). A sequence is a function whose inputs are the positive integers \(1, 2, 3, \dots\)
There are two common ways to define a sequence.
- An explicit formula gives \(a_n\) directly from \(n\). For example, \(a_n = n^2 - 1\) gives \(0, 3, 8, 15, \dots\), and you can jump straight to \(a_{50}\).
- A recursive formula gives the first term and a rule that builds each term from the previous one, such as \(a_1 = 4\) and \(a_{n+1} = a_n + 3\), which gives \(4, 7, 10, 13, \dots\)
2. Arithmetic sequences
A sequence is arithmetic if the difference between consecutive terms is always the same number \(d\), the common difference: \(a_{n+1} - a_n = d\).
\[ a_n = a_1 + (n-1)d \]
Recursive form: \(a_{n+1} = a_n + d\).
The points of an arithmetic sequence lie on a straight line, and the common difference is its slope.
Consider \(7, 12, 17, 22, \dots\) The common difference is \(d = 12 - 7 = 5\), so \(a_n = 7 + (n-1)\cdot 5 = 5n + 2\). The 20th term is \(a_{20} = 5\cdot 20 + 2 = 102\).
3. Geometric sequences
A sequence is geometric if each term is the previous one multiplied by the same nonzero number \(r\), the common ratio: \(\dfrac{a_{n+1}}{a_n} = r\).
\[ a_n = a_1 \cdot r^{\,n-1} \]
Recursive form: \(a_{n+1} = r \cdot a_n\).
For \(3, 6, 12, 24, \dots\) the ratio is \(r = 6 \div 3 = 2\), so \(a_n = 3\cdot 2^{n-1}\). Then \(a_8 = 3\cdot 2^7 = 3\cdot 128 = 384\).
The exponent is \(n-1\), not \(n\). Check with \(n = 1\): you must get \(a_1\) back. Also, a ratio between \(0\) and \(1\) makes the terms shrink, and a negative ratio makes them alternate in sign.
4. Explicit and recursive formulas
- Decide whether the pattern is arithmetic (constant difference) or geometric (constant ratio).
- Find \(a_1\) and \(d\) (or \(r\)).
- Recursive form: write \(a_1\) and \(a_{n+1} = a_n + d\) (or \(a_{n+1} = r\,a_n\)).
- Explicit form: use \(a_n = a_1 + (n-1)d\) (or \(a_n = a_1 r^{n-1}\)).
- Test both forms on the first three terms.
If you know two terms of an arithmetic sequence, subtract them and divide by the number of steps between them to get \(d\). For a geometric sequence, divide them and take the root of the same order to get \(r\).
5. Arithmetic series
A series is the sum of the terms of a sequence. The partial sum \(S_n = a_1 + a_2 + \dots + a_n\) is the sum of the first \(n\) terms.
To add \(1 + 2 + 3 + 4 + 5\), put a copy of the same dots upside down next to the first one. You get a rectangle of 5 rows and 6 columns, so the double of the sum is \(5 \cdot 6\).
\[ S_n = \dfrac{n\,(a_1 + a_n)}{2} \qquad\text{or}\qquad S_n = \dfrac{n\,[\,2a_1 + (n-1)d\,]}{2} \]
In words: the number of terms times the average of the first and last terms.
Add \(5 + 9 + 13 + \dots\) up to 15 terms. Here \(d = 4\) and \(a_{15} = 5 + 14\cdot 4 = 61\). Then \(S_{15} = \dfrac{15\,(5 + 61)}{2} = \dfrac{15\cdot 66}{2} = 495\).
6. Finite geometric series
For \(r \neq 1\):
\[ S_n = a_1\,\dfrac{1 - r^{\,n}}{1 - r} \]
If \(r = 1\), every term equals \(a_1\) and \(S_n = n\,a_1\).
Why it works. Multiply \(S_n = a_1 + a_1 r + \dots + a_1 r^{n-1}\) by \(r\) and subtract: \(S_n - rS_n = a_1 - a_1 r^n\). Almost every term cancels, and dividing by \(1-r\) gives the formula.
A game awards 2 points for the first level, and each level pays 3 times as many points as the one before. Total for 6 levels: \(a_1 = 2\), \(r = 3\), so \(S_6 = 2\cdot\dfrac{1 - 3^6}{1 - 3} = 2\cdot\dfrac{-728}{-2} = 728\) points.
7. Infinite geometric series
Can you add infinitely many numbers and get a finite answer? Yes, when the terms shrink fast enough. Cut a square of area 1 in half, then cut half of what remains, and so on: the pieces \(\tfrac12 + \tfrac14 + \tfrac18 + \dots\) fill the square, so their sum is 1.
If \(|r| < 1\), the series converges and
\[ S = \dfrac{a_1}{1 - r} \]
If \(|r| \ge 1\), the series diverges: it has no finite sum.
The reason is that \(r^n\) gets closer and closer to 0 when \(|r| < 1\), so the partial sums \(a_1\dfrac{1 - r^n}{1 - r}\) approach \(\dfrac{a_1}{1-r}\). The chart shows the partial sums of \(8 + 4 + 2 + 1 + \dots\) closing in on \(\dfrac{8}{1 - 1/2} = 16\).
For \(12 + 4 + \dfrac43 + \dots\) we have \(r = \dfrac{4}{12} = \dfrac13\), and \(|r| < 1\). So \(S = \dfrac{12}{1 - \frac13} = \dfrac{12}{\frac23} = 18\).
On my home planet we test convergence first: look at the ratio. If it is not strictly between \(-1\) and \(1\), stop, because there is no sum to compute!
8. Sigma notation
\[ \sum_{k=m}^{n} a_k = a_m + a_{m+1} + \dots + a_n \]
\(k\) is the index, \(m\) is the lower limit, and \(n\) is the upper limit. The number of terms is \(n - m + 1\).
\(\displaystyle\sum_{k=1}^{5}(3k - 1) = 2 + 5 + 8 + 11 + 14 = 40\). It is arithmetic with \(a_1 = 2\), \(a_5 = 14\), so the formula gives \(\dfrac{5\,(2 + 14)}{2} = 40\). The same sum written for a geometric series is \(\displaystyle\sum_{k=1}^{\infty} 5\left(\tfrac12\right)^{k-1} = \dfrac{5}{1 - \frac12} = 10\).
When the lower limit is not 1, count the terms with \(n - m + 1\). For instance, \(\sum_{k=3}^{9}\) has 7 terms, not 6 or 9.
Useful rules: \(\sum (a_k + b_k) = \sum a_k + \sum b_k\) and \(\sum c\,a_k = c\sum a_k\).
Key takeaways
- Arithmetic: constant difference \(d\), \(a_n = a_1 + (n-1)d\).
- Geometric: constant ratio \(r\), \(a_n = a_1 r^{n-1}\).
- A recursive formula needs a starting term and a rule; an explicit formula gives \(a_n\) directly.
- Arithmetic series: \(S_n = \dfrac{n(a_1 + a_n)}{2}\).
- Finite geometric series: \(S_n = a_1\dfrac{1 - r^n}{1 - r}\) for \(r \ne 1\).
- Infinite geometric series: \(S = \dfrac{a_1}{1-r}\) only when \(|r| < 1\).
- In sigma notation, the number of terms is upper limit minus lower limit plus 1.
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