# Sequences

A-Level Maths revision looking at Sequences including Notation, Convergent Sequences and Recurrence Relations.

**nth Term**

In the sequence 2, 4, 6, 8, 10... there is an obvious pattern. Such sequences can be expressed in terms of the nth term of the sequence. In this case, the nth term = 2n. To find the 1st term, put n = 1 into the formula, to find the 4th term, replace the n"s by 4"s: 4th term = 2 × 4

= 8.

**Trial and Error**

**Example**

What is the nth term of the sequence 2, 5, 10, 17, 26... ?

Let"s use trial and error (essentially guessing what we think will work):

n | = | 1 | 2 | 3 | 4 | 5 |

n² | = | 1 | 4 | 9 | 16 | 25 |

n² + 1 | = | 2 | 5 | 10 | 17 | 26 |

This is the required sequence, so the nth term is n² + 1. For some sequences, there is no easy way of working out the nth term of a sequence, other than to try different possibilities.

Tips: if the sequence is going up in threes (e.g. 3, 6, 9, 12...), there will probably be a three in the formula, etc.

In many cases, square numbers will come up, so try squaring n, as above. Also, the triangular numbers formula often comes up. This is ½n(n + 1).

**Notation**

The nth term of a sequence is sometimes written as U_{n} . So in the last example, U_{n} = n² + 1 . The 5th term is therefore U_{5} = 25 + 1 = 26.

**Convergent Sequences**

Sequences whose nth term approaches a finite number as n becomes larger are known as convergent sequences and the number to which the sequence converges is known as the limit of the sequence. For example: 10, 5, 2.5, 1.25, 0.625, ... converges (gets closer and closer) towards the limit zero.

**Recurrence Relations**

This is where the next term of a sequence is defined using the previous term(s). To define a recurrence relation, you have to give the first term. Then you give a formula to tell you how to work out the next term from the previous ones.

For example, consider the sequence: 2, 4, 8, 16, 32, ... . Each term in the sequence is got by doubling the previous term. So to define the recurrence relation, we give the first term, written U_{1} = 2. Then we write: U_{n} = 2(U_{n-1}). This just means that the nth term, U_{n} is equal to 2 × the (n-1)th term, U_{n-1}.