Sequences & Series

A sequence is a set of numbers or algebraic expressions defined by a rule.

A term is an element of a sequence — it is one part of the sequence.

Individual terms are denoted by the letter T or U with a subscript.

T_n is called the general term. If we know T_n we can find any term in the sequence. If we are given the terms, we can find T_n.

T_n+1 is the next term up from T_n. And n ∈ ℕ — n must be a whole positive number.

Heads up — two notations.

U_n = T_n

Both letters mean the same thing. Exams use T_n and U_n interchangeably. When you see one in a question, you can use the other in your working.

T₁, T₂, T₃, … , T_n−1, T_n, T_n+1, …

A series is the sum of the terms of a sequence.

S₁ = U₁

S₂ = U₁ + U₂

S₃ = U₁ + U₂ + U₃

From each sequence we can construct a series S_n:

S_n = U₁ + U₂ + … + U_n

The series S_n can be written as:

S_n = Σⁿ_n=1 U_n   or   S_n = Σⁿ_r=1 U_r

where r ∈ ℕ and r ≤ n.

Three pieces to remember: where to start (the bottom of the Σ), where to end (the top), and the rule for the term (after the Σ).

U_n = S_n − S_n−1

Get S_n−1 by replacing every n in S_n with (n−1), then subtract.

The free check. Verify at n = 1: since S₁ = T₁, your two formulas must agree at n = 1. If they don't, you dropped a minus sign in the subtraction — go back and re-distribute.

A sequence which obeys the rule U_n − U_n−1 = constant is called an arithmetic sequence.

The corresponding series is an arithmetic series (or arithmetic progression). Same idea as a linear sequence — same gap each step.

The constant is called the common difference and is usually denoted by d = U₂ − U₁.

The first term is usually denoted by a = U₁.

Also note: a = T₁ = S₁ — the first term and the first partial sum are the same thing. Useful when you're given S_n and need a.

T_n = a + (n − 1)d In Tables

It's in the Log Tables — you don't need to memorise it.

If you can see the same gap each step, you can write T_n directly — no need for a + (n − 1)d.

The rule looks like y = mx + c:

T_n = (gap) · n + (constant)

The gap is the slope (like m); the constant is whatever you need to add to pin it to the right starting value.

e.g. 5, 8, 11, 14, … — gap is 3, starting value is 5, so T_n = 3n + 2 (check: T₁ = 3 + 2 = 5 ✓).

S_n = n2 { 2a + (n − 1)d } In Tables

Also in the Log Tables — you don't need to memorise it.

Use simultaneous equations.

Use the idea that

T₃ − T₂ = T₂ − T₁

e.g. the first three terms are x − 3, 2x + 7, x + 5 — find x.

Let the three terms be

x − y,   x,   x + y

e.g. three consecutive AP terms add to 12 and multiply to 48 — find them.

A sequence which obeys the rule

T_n+1T_n = constant

is called a geometric sequence.

To get from one term to the next we multiply (or divide) by the same number.

The number we multiply by is called the common ratio, usually denoted by r, where

r = T₂T₁   =   T_n+1T_n

The general form T_n+1 / T_n is useful when you check a sequence is geometric — the ratio between any two consecutive terms must be the same.

First term: a = T₁ = S₁.

The value of r tells you what the sequence does:

r > 1    terms grow without bound

0 < r < 1    terms shrink towards zero

r < 0    terms alternate sign as they go

T_n = arⁿ⁻¹ In Tables

It's in the Log Tables — you don't need to memorise it.

Compound interest is geometric — each year multiplies last year's balance by the same factor (1 + i). The formula is:

F = P(1 + i)^t In Tables

P = present value (principal)
F = future value
i = interest rate as a decimal
t = time

e.g. €300 saved at 7% for one year gives F = 300(1.07); after t years F = 300(1.07)^t.

S_n = a(1 − rⁿ)1 − r In Tables

Use simultaneous equations.

Use the idea that

T₂T₁ = T₃T₂

e.g. 2, x + 1, 32 are the first three terms in a GP — find x.

Let the three terms be

xy,   x,   xy

e.g. three consecutive GP terms add to 14 and multiply to 64 — find them.

The infinite geometric series

a + ar + ar² + …

is convergent if |r| < 1, in which case

S_∞ = lim
n→∞ S_n = a1 − r In Tables

|r| < 1 means that r is a proper fraction.

Each term in a sequence depends on the terms before it.

U_n+1 depends on U_n.

You always need a starting value (or two, for a two-back recurrence).

Given a closed form U_n and asked to prove it satisfies a recurrence.

The trick: when

U_n = pⁿ + qⁿ

let

a = pⁿ,   b = qⁿ

so U_n = a + b. Then work out U_n+1 and U_n+2 in terms of a and b using the index law a^p+q = a^p · a^q.

Sub everything into the recurrence — the coefficients of a and b will both collapse to zero.

Given U_n is the nth term:

(i) If U_n+1 > U_n for all n, the sequence is increasing.

(ii) If U_n+1 < U_n for all n, the sequence is decreasing.

For positive terms:

U_n+1U_n > 1   ⟹   increasing

U_n+1U_n < 1   ⟹   decreasing

When the gap of the gap is the same, use

T_n = an² + bn + c

where

a = gap of gap2

A series of the form

U₁ + U₂ + U₃ + … + U_∞ = Σ^∞_n=1 U_n

S_∞ = Σ^∞_n=1 U_n = lim
n→∞ S_n

An arithmetic series can never have a limit and is therefore always divergent.

S_∞ = lim
n→∞ S_n

Step 1. Find a concise expression for S_n.

Step 2. Evaluate the limit of S_n as n approaches infinity.

To evaluate the limit, use the rules from differentiation.

lim
n→∞ rⁿ = 0   given that |r| < 1.