A savings plan that adds the same amount every month, a rumor that doubles in reach every day, a ball that loses height on every bounce: all of them are described by lists of numbers that follow a rule. In this chapter you will learn to write that rule, add up long lists quickly, decide when an infinite list has a finite total, and prove statements that hold for every whole number. You will finish with a powerful shortcut for expanding powers of a binomial.
1. Sequences and notation
A sequence is an ordered list of numbers \(a_1, a_2, a_3, \dots\). The number \(a_n\) is the \(n\)th term, and \(n\) is its index. A sequence can be defined explicitly, by a formula for \(a_n\) in terms of \(n\), or recursively, by giving the first term(s) and a rule that builds each term from the previous ones.
For instance, \(a_n = n^2 + 1\) is explicit: \(a_4 = 17\) immediately. By contrast, \(a_1 = 3\) and \(a_n = a_{n-1} + 5\) is recursive: you must find \(a_2 = 8\), then \(a_3 = 13\), and so on. A series is the sum of the terms of a sequence. The sum of the first \(n\) terms is called the partial sum and is written \(S_n = a_1 + a_2 + \dots + a_n\).
2. Arithmetic sequences and series
A sequence is arithmetic if consecutive terms differ by a constant \(d\), called the common difference: \(a_{n+1} - a_n = d\) for every \(n\).
\[ a_n = a_1 + (n-1)d \qquad\text{and}\qquad S_n = \dfrac{n\,(a_1 + a_n)}{2} = \dfrac{n\,\bigl(2a_1 + (n-1)d\bigr)}{2}. \]
The sum formula comes from writing the sum forwards and backwards. Each of the \(n\) vertical pairs adds up to \(a_1 + a_n\), so twice the sum is \(n(a_1 + a_n)\). The picture below shows the sequence 5, 8, 11, 14, ... : every bar is 3 units taller than the previous one.
An arithmetic sequence has \(a_1 = 7\) and \(d = 4\). Find \(a_{20}\) and \(S_{20}\).
\(a_{20} = 7 + 19 \times 4 = 83\). Then \(S_{20} = \dfrac{20\,(7 + 83)}{2} = 10 \times 90 = 900\).
3. Geometric sequences and series
A sequence is geometric if each term is the previous one multiplied by a constant \(r \neq 0\), the common ratio: \(a_{n+1} = r\,a_n\).
\[ a_n = a_1\, r^{\,n-1} \qquad\text{and, for } r \neq 1,\qquad S_n = a_1\,\dfrac{1 - r^{\,n}}{1 - r}. \]
If \(r = 1\), every term equals \(a_1\) and \(S_n = n\,a_1\).
To see where the sum formula comes from, compute \(S_n - rS_n\): almost everything cancels and you are left with \(a_1 - a_1 r^n\), so \(S_n(1-r) = a_1(1 - r^n)\). The bars below show 2, 6, 18, 54, 162 (ratio 3): geometric growth quickly outruns arithmetic growth.
Find the sum of the first 6 terms of the geometric sequence 5, 15, 45, ...
Here \(a_1 = 5\) and \(r = 3\). So \(S_6 = 5\cdot\dfrac{3^6 - 1}{3 - 1} = 5 \cdot \dfrac{728}{2} = 1820\). Check by adding: \(5 + 15 + 45 + 135 + 405 + 1215 = 1820\).
4. Infinite geometric series
What happens to \(S_n = a_1\dfrac{1 - r^n}{1 - r}\) when \(n\) gets huge? If \(|r| \lt 1\), then \(r^n\) shrinks toward 0, so \(S_n\) settles down to a single number. If \(|r| \ge 1\), the terms do not shrink and the partial sums do not settle.
If \(|r| \lt 1\), the series \(a_1 + a_1 r + a_1 r^2 + \cdots\) converges and its sum is \[ S = \dfrac{a_1}{1 - r}. \]
If \(|r| \ge 1\) (and \(a_1 \neq 0\)), the series diverges: it has no finite sum.
The graph shows the partial sums of \(1 + \tfrac12 + \tfrac14 + \cdots\). The dots climb toward the dashed line \(y = 2\), which is exactly \(\dfrac{1}{1 - 1/2}\).
Evaluate \(12 - 4 + \dfrac{4}{3} - \cdots\)
The first term is 12 and the ratio is \(r = -\dfrac{4}{12} = -\dfrac13\). Since \(|r| \lt 1\), the series converges: \(S = \dfrac{12}{1 - (-1/3)} = \dfrac{12}{4/3} = 9\).
Never use \(\dfrac{a_1}{1-r}\) without checking \(|r| \lt 1\) first. For \(3 + 6 + 12 + \cdots\) the formula would give \(-3\), which is nonsense for a sum of positive numbers.
5. Sigma notation
The symbol \(\displaystyle\sum_{k=m}^{n} a_k\) means \(a_m + a_{m+1} + \cdots + a_n\). The letter \(k\) is the index of summation, \(m\) is the lower limit and \(n\) the upper limit.
\[ \sum_{k=1}^{n} (a_k + b_k) = \sum_{k=1}^{n} a_k + \sum_{k=1}^{n} b_k, \qquad \sum_{k=1}^{n} c\,a_k = c\sum_{k=1}^{n} a_k, \qquad \sum_{k=1}^{n} c = nc. \]
Useful formulas: \(\displaystyle\sum_{k=1}^{n} k = \dfrac{n(n+1)}{2}\) and \(\displaystyle\sum_{k=1}^{n} k^2 = \dfrac{n(n+1)(2n+1)}{6}\).
Evaluate \(\displaystyle\sum_{k=1}^{5} (2k+1)\) two ways.
Expanding: \(3 + 5 + 7 + 9 + 11 = 35\). With the rules: \(2\cdot\dfrac{5\cdot 6}{2} + 5 = 30 + 5 = 35\).
6. Recursive definitions
A recursive definition needs two ingredients: initial term(s) and a recurrence relation. An arithmetic sequence is \(a_n = a_{n-1} + d\), a geometric one is \(a_n = r\,a_{n-1}\). Many sequences are neither, for example the Fibonacci-style rule \(f_n = f_{n-1} + f_{n-2}\), which needs two starting values.
- Compute the first four or five terms with the recurrence.
- Look for a pattern (constant differences, constant ratios, powers, squares).
- Write a conjecture for \(a_n\).
- Prove it by induction (next section).
Let \(a_1 = 3\) and \(a_n = 2a_{n-1} - 1\). The terms are 3, 5, 9, 17. Each is one more than a power of 2, so we conjecture \(a_n = 2^n + 1\). Check the recurrence: \(2(2^{n-1} + 1) - 1 = 2^n + 1\). The formula is consistent.
7. Mathematical induction
Suppose you want to prove that a statement \(P(n)\) is true for every integer \(n \ge 1\). You cannot test infinitely many cases, but you can use the domino idea shown below: if the first domino falls, and each falling domino knocks over the next, then they all fall.
Let \(P(n)\) be a statement about an integer \(n\). If
- Base case: \(P(1)\) is true, and
- Inductive step: for every \(k \ge 1\), if \(P(k)\) is true (the inductive hypothesis), then \(P(k+1)\) is true,
then \(P(n)\) is true for all integers \(n \ge 1\).
Prove that \(1 + 3 + 5 + \cdots + (2n-1) = n^2\) for all \(n \ge 1\).
Base case. For \(n = 1\): the left side is 1 and the right side is \(1^2 = 1\).
Inductive step. Assume \(1 + 3 + \cdots + (2k-1) = k^2\). Then \(1 + 3 + \cdots + (2k-1) + (2k+1) = k^2 + 2k + 1 = (k+1)^2\), which is the statement for \(n = k+1\).
By induction, the formula holds for every \(n \ge 1\).
On my home planet we say: the inductive step must use the hypothesis. If your proof of \(P(k+1)\) never mentions \(P(k)\), something is missing!
8. The Binomial Theorem
Expanding \((a+b)^n\) by hand is slow, but the coefficients follow a pattern. They are the binomial coefficients \(\dbinom{n}{k} = \dfrac{n!}{k!\,(n-k)!}\), the number of ways to choose \(k\) objects from \(n\). They form Pascal’s triangle: each entry is the sum of the two entries above it.
For every integer \(n \ge 0\), \[ (a+b)^n = \sum_{k=0}^{n} \dbinom{n}{k}\, a^{\,n-k}\, b^{\,k}. \]
The expansion has \(n+1\) terms, and the coefficients add up to \(2^n\).
Expand \((x+2)^4\).
Row 4 of Pascal’s triangle is 1, 4, 6, 4, 1. So \((x+2)^4 = x^4 + 4(2)x^3 + 6(4)x^2 + 4(8)x + 16 = x^4 + 8x^3 + 24x^2 + 32x + 16\).
Do not forget to raise the second term to the power too, including its coefficient and its sign: in \((x-2)^4\) the powers of \(-2\) alternate in sign.
Key takeaways
- Arithmetic: \(a_n = a_1 + (n-1)d\) and \(S_n = \dfrac{n(a_1 + a_n)}{2}\).
- Geometric: \(a_n = a_1 r^{n-1}\) and \(S_n = a_1\dfrac{1-r^n}{1-r}\) for \(r \neq 1\).
- An infinite geometric series converges only if \(|r| \lt 1\), and then \(S = \dfrac{a_1}{1-r}\).
- Sigma notation packs a sum into one line, and sums split over addition and constant multiples.
- A recursive definition needs a starting value and a rule; conjecture a formula, then prove it.
- Induction: prove the base case, then prove that \(P(k)\) implies \(P(k+1)\).
- Binomial Theorem: \((a+b)^n = \sum_{k=0}^{n}\dbinom{n}{k}a^{n-k}b^k\), with coefficients from Pascal’s triangle.
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