Sequences and Series

Updated 7 October 2026

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?

An nth-term formula gives any term of a sequence from its position n. Put in n = 100 and it gives the hundredth term, while a term-to-term rule such as "add 3" needs 99 steps to get there. See Term-to-Term Rules. For an arithmetic sequence, the common difference is the coefficient of n. In 7, 11, 15, 19 the difference is 4, so start from 4n, which gives 4, 8, 12, 16; every term is 3 more, so the nth term is 4n + 3. This works because the first term is 4 × 1 + 3 = 7, and each step adds one more 4. Only the constant depends on the first term, and subtracting finds it. The method fails when the differences change, as in 1, 4, 9, 16, whose nth term is $n^2$n². Check the formula on a term you did not use: the 4th term is 4 × 4 + 3 = 19.

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. 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?

An arithmetic series is added by pairing its 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, so n terms from a to l add to $\frac{n\left(a+l\right)}{2}$n(a + l)/2. In 3 + 7 + 11 + … + 39 the terms rise by 4, so there are (39 − 3) ÷ 4 + 1 = 10 of them. Each pair totals 3 + 39 = 42, and five pairs give 5 × 42 = 210. The pairs match because moving inward adds 4 at one end and removes 4 at the other. With an odd number of terms, the unpaired middle term counts as half a pair. When the last term is unknown, use $\frac{n\left(2a+\left(n-1\right)d\right)}{2}$n(2a + (n − 1)d)/2 with the difference d. Confirm with the mean term, (3 + 39) ÷ 2 = 21: 10 × 21 = 210.

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?

An infinite sum can be finite because its terms can shrink fast enough for the running totals to close in on a limit, the sum to infinity. 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 half of the gap left to 1, so after n terms the gap is $\frac{1}{2^n}$1/2ⁿ, and no total passes 1. A geometric series with first term a and ratio r has sum to infinity $\frac{a}{\left(1-r\right)}$a/(1 − r), valid only when r is strictly between −1 and 1. Here $a=\frac{1}{2}$a = 1/2 and $1-r=\frac{1}{2}$1 − r = 1/2, so the sum is $\frac{1}{2}÷\frac{1}{2}=1$1/2 ÷ 1/2 = 1. The formula is the limit of the finite sum $\frac{a\left(1-r^n\right)}{\left(1-r\right)}$a(1 − rⁿ)/(1 − r), as $r^n$rⁿ shrinks to 0. Shrinking terms alone are not enough: $1+\frac{1}{2}+\frac{1}{3}+\dots$1 + 1/2 + 1/3 + … grows past every bound. Check a partial total: $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}=\frac{7}{8}$1/2 + 1/4 + 1/8 = 7/8, which is $\frac{1}{8}$1/8 short of 1.

By that formula, $\frac{a}{\left(1-r\right)}$a/(1 − r), the series 8 + 4 + 2 + 1 + … with a = 8 and $r=\frac{1}{2}$r = 1/2 has the sum $\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?

Induction proves a formula for every whole number n in two steps. The base case shows the statement is true for n = 1. The inductive step assumes it is true for n = k and shows it must then be true for n = k + 1. Together the two steps reach every n. To prove that the first n odd numbers add to $n^2$n², check n = 1: $1=1^2$1 = 1². Assume the first k odd numbers add to $k^2$k², then add the next, 2k + 1: $k^2+2k+1={\left(k+1\right)}^2$k² + 2k + 1 = (k + 1)², the statement for k + 1. This is not circular: it assumes one case to prove the next. Both steps are needed: $n^2+n+41$n² + n + 41 is prime for every n from 1 to 39, yet no inductive step can be proved, and n = 40 fails. Check one case directly: $1+3+5+7=16=4^2$1 + 3 + 5 + 7 = 16 = 4².

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 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?

A series converges when its partial sums approach a limit. Start with the cheapest test, the nth term 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. See The nth Term Test for Divergence. The test works because a convergent series must eventually add almost nothing. It only ever proves divergence: terms tending to zero are necessary for convergence, not sufficient, and the harmonic series $1+\frac{1}{2}+\frac{1}{3}+\dots$1 + 1/2 + 1/3 + … diverges although its terms shrink. When the terms do tend to zero, choose a sharper test by the shape of the term: a p-series such as $\sum \frac{1}{n^2}$Σ 1/n² converges because its power exceeds 1, and the ratio test handles factorials. Check the example by computing terms: the 99th is $\frac{99}{199}$99/199, close to a half, and the first two add to $\frac{1}{3}+\frac{2}{5}=\frac{11}{15}$1/3 + 2/5 = 11/15.

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.

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. Related guides: Applications of Integration and Polar Curves and Complex Numbers.