Sequences and Series

A sequence is a list of numbers made by a rule. A series is the sum of its terms. The sequence 1, 2, 3, 4 has four terms; the series 1 + 2 + 3 + 4 is 10. A sequence question asks for a term. A series question asks for a total.

What is the nth term, and how do you find it?

The nth term is a formula that gives any term from its position: put in n = 100 and it gives the hundredth term. A term-to-term rule such as "add 3" needs 499 steps to reach term 500. See Term-to-Term Rules.

When a sequence rises by the same amount each time, that amount is the common difference, and it is the coefficient of n. In 5, 8, 11, 14 the common difference is 3, so start from 3n, which gives 3, 6, 9, 12. Every term is 2 more than that, so the nth term is 3n + 2. The first term only fixes the constant, and you find the constant by subtracting. See The nth Term.

Some sequences are recognized rather than worked out: the squares 1, 4, 9, 16, the triangle numbers 1, 3, 6, 10, the powers of two 1, 2, 4, 8, and Fibonacci 1, 1, 2, 3, 5, 8, where each term is the sum of the two before it. See Sequences Worth Knowing.

When the differences are not constant

If the first differences are not constant but the second differences are, the sequence is quadratic and its nth term contains $n^2$n². The coefficient of $n^2$n² is half the second difference.

Here the second difference is 2, so the coefficient of $n^2$n² is 1. Subtract $n^2$n² from each term and 2, 4, 6, 8 is left, which is 2n. The nth term is $n^2+2n$n² + 2n. See Quadratic Sequences.

If instead each term is a fixed multiple of the one before, the sequence is geometric and that multiple is the common ratio. For 3, 6, 12, 24, with ratio 2, the nth term is $3\times 2^{n-1}$3 × 2ⁿ⁻¹. The exponent is n − 1 because the first term has been multiplied by the ratio zero times. See Geometric Sequences.

A recurrence relation writes the term-to-term rule in symbols, with a first term: $u_{n+1}=u_n+4$uₙ₊₁ = uₙ + 4 and $u_1=3$u₁ = 3 gives 3, 7, 11, 15. Without the first term the rule names no sequence. See Recurrence Relations.

How do you add up a long list?

To add an arithmetic series, pair the terms from the outside in: first with last, second with second-last. Every pair has the same total, and there are half as many pairs as terms.

Sigma notation writes a sum in one line. The letter under the $\sum$Σ is the counter, the numbers below and above it are where the counter starts and stops, and the expression to the right is the term to add. $\sum _{n=1}^{4}2n$Σ 2n from n = 1 to 4 is 2 + 4 + 6 + 8 = 20. See Sigma Notation.

A geometric series needs a different move. Write S = 3 + 6 + 12 + 24, multiply by the common ratio to get 2S = 6 + 12 + 24 + 48, and subtract the first line from the second, leaving S = 48 − 3 = 45. In letters the same subtraction gives $S=\frac{a\left(r^n-1\right)}{\left(r-1\right)}$S = a(rⁿ − 1)/(r − 1). See The Sum of a Geometric Series.

Why can an infinite sum be finite?

Because the terms can shrink fast enough that the running total closes in on a limit. Add $\frac{1}{2}$1/2, then $\frac{1}{4}$1/4, then $\frac{1}{8}$1/8: the totals are 0.5, 0.75, 0.875. Each term covers exactly half of the gap that is left, so the total never passes 1, and the sum to infinity is that limit, 1.

For a geometric series with first term a and common ratio r, the sum to infinity is $\frac{a}{\left(1-r\right)}$a/(1 − r), and it exists only when r is strictly between −1 and 1. 8 + 4 + 2 + 1 + … has a = 8 and $r=\frac{1}{2}$r = 1/2, so its sum is $\frac{8}{\left(1-\frac{1}{2}\right)}=16$8/(1 − 1/2) = 16. See The Sum to Infinity.

How do you prove a formula holds for every n?

By induction. Prove the statement for n = 1, then prove that if it holds for one value of n it holds for n + 1. The first case is true, so the second is, so the third is, and the chain never stops.

Take $1+2+\cdots +n=\frac{n\left(n+1\right)}{2}$1 + 2 + … + n = n(n + 1)/2. At n = 1 both sides are 1, and adding n + 1 to both sides turns the case for n into the case for n + 1. See Proof by Induction: Summing 1 to n.

The inductive step assumes one instance of the statement to prove the next. It never assumes the statement itself, so it is not circular. The same two steps prove that $8^n-1$8ⁿ − 1 is a multiple of 7 for every n: check n = 1, then show that moving to the next case adds another multiple of 7. See Proof by Induction: Divisibility.

The first n squares and the first n cubes also have closed formulas; the cubes add to ${\left(1+2+\cdots +n\right)}^2$(1 + 2 + … + n)². See Sums of Squares and Cubes. The method of differences writes each term as a difference of two consecutive values: $\frac{1}{\left(r\left(r+1\right)\right)}$1/(r(r + 1)) is $\frac{1}{r}-\frac{1}{\left(r+1\right)}$1/r − 1/(r + 1), so the sum from r = 1 to n collapses to $1-\frac{1}{\left(n+1\right)}$1 − 1/(n + 1). See The Method of Differences.

Does this series converge?

Start with the cheapest test. If the terms do not tend to zero, the series diverges. The terms of $\sum \frac{n}{\left(2n+1\right)}$Σ n/(2n + 1) tend to $\frac{1}{2}$1/2, so that series diverges. This test only ever proves divergence: terms tending to zero are necessary for convergence, not sufficient. See The nth Term Test for Divergence.

The p-series $\sum \frac{1}{n^p}$Σ 1/nᵖ converges exactly when p > 1, so $\sum \frac{1}{n^2}$Σ 1/n² converges and the harmonic series $\sum \frac{1}{n}$Σ 1/n, where p = 1, does not. The test compares the sum with the area under $y=\frac{1}{x^p}$y = 1/xᵖ. See The p-Series.

A comparison test uses a series you already know. A positive series whose terms stay below those of a convergent series converges too; one whose terms stay above those of a divergent series diverges too. $\sum \frac{1}{\left(n^2+1\right)}$Σ 1/(n² + 1) converges, because each term is below $\frac{1}{n^2}$1/n². See Comparison Tests for Series.

The ratio test handles factorials and powers. Let L be the limit of each term divided by the one before: L < 1 converges, L > 1 diverges, and L = 1 decides nothing. For $\sum \frac{2^n}{n!}$Σ 2ⁿ/n! the ratio is $\frac{2}{\left(n+1\right)}$2/(n + 1), which tends to 0, so the series converges. See The Ratio Test.

When the signs alternate the conditions are lighter: sizes that decrease and tend to zero are enough, because the limit is trapped between consecutive partial sums. So $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots$1 − 1/2 + 1/3 − 1/4 + … converges while the harmonic series does not. See The Alternating Series Test.

Turning a function into an infinite polynomial

A power series has terms containing powers of x, so it converges for some x and not others. The ratio test's condition L < 1 says x must lie within a distance R of the center, and R is the radius of convergence. The test says nothing at the two endpoints, so check each separately. For $\sum \frac{x^n}{n}$Σ xⁿ/n the radius is 1: at x = 1 it is the harmonic series and diverges, at x = −1 it alternates and converges, so the interval is $-1\le x<1$−1 ≤ x < 1. See Radius and Interval of Convergence.

A Maclaurin series writes a function as a power series centered at 0. The nth coefficient is the nth derivative at 0 divided by n!, so the series and the function agree in value and in every derivative at 0. Know $e^x$eˣ, sin x, cos x, ln(1 + x) and $\frac{1}{\left(1-x\right)}$1/(1 − x); the rest come from these by substituting, differentiating or integrating term by term. See The Standard Maclaurin Series.

A Taylor series uses a different center a: derivatives at a, and powers of x − a. ln x has no value at 0, so its series is built at x = 1. See Taylor Series About a Point.

Cutting a series after degree n leaves an error. The Lagrange error bound says it is at most $M\times |x-\frac{a{|}^{n+1}}{\left(n+1\right)!}$M × |x − a|ⁿ⁺¹/(n + 1)!, where M is the largest size of the (n + 1)th derivative between a and x. For sin x to degree 3 on $|x|\le 1$|x| ≤ 1, every derivative has size at most 1, so the error is at most $\frac{1}{24}$1/24. A bound is the most the error can be, not the error itself. See The Lagrange Error Bound.

The mistakes worth naming

• Using the first term as the coefficient of n. The coefficient of n is the common difference. The first term only fixes the constant.
• Writing arⁿ for the nth term of a geometric sequence. It is arⁿ⁻¹, because the first term has been multiplied by r zero times.
• Reading the nth term test backwards. Terms tending to zero do not prove convergence.
• Using $\frac{a}{\left(1-r\right)}$a/(1 − r) when $|r|\ge 1$|r| ≥ 1. The formula returns a number that means nothing, because the series has no sum.

Learn this properly in the app

Every lesson named on this page is in Math Challenge, with its figures and practice questions. The neighboring guides are quadratics and polynomials, where the binomial theorem produces series of its own, and functions and rational functions, which sets up the exponential and the logarithm used in the Maclaurin list.