### 1 Introduction

A sequence is any succession of numbers. For example the sequence

$\phantom{\rule{2em}{0ex}}1,\phantom{\rule{1em}{0ex}}1,\phantom{\rule{1em}{0ex}}2,\phantom{\rule{1em}{0ex}}3,\phantom{\rule{1em}{0ex}}5,\phantom{\rule{1em}{0ex}}8,\phantom{\rule{1em}{0ex}}\dots$

which is known as the Fibonacci sequence, is formed by adding two consecutive terms together to obtain the next term. The numbers in this sequence continually increase without bound and we say this sequence diverges . An example of a convergent sequence is the harmonic sequence

$\phantom{\rule{2em}{0ex}}1,\phantom{\rule{1em}{0ex}}\frac{1}{2},\phantom{\rule{1em}{0ex}}\frac{1}{3},\phantom{\rule{1em}{0ex}}\frac{1}{4},\phantom{\rule{1em}{0ex}}\dots$

Here we see the magnitude of these numbers continually decrease and it is obvious that the sequence converges to the number zero. The related alternating harmonic sequence

$\phantom{\rule{2em}{0ex}}1,\phantom{\rule{1em}{0ex}}-\frac{1}{2},\phantom{\rule{1em}{0ex}}\frac{1}{3},\phantom{\rule{1em}{0ex}}-\frac{1}{4},\phantom{\rule{1em}{0ex}}\dots$

is also convergent to the number zero. Whether or not a sequence is convergent is often easy to deduce by graphing the individual terms. The diagrams in Figure 1 show how the individual terms of the harmonic and alternating harmonic series behave as the number of terms increase.

Figure 1

Graph the sequence:

$\phantom{\rule{2em}{0ex}}1,\phantom{\rule{1em}{0ex}}-1,\phantom{\rule{1em}{0ex}}1,\phantom{\rule{1em}{0ex}}-1,\phantom{\rule{1em}{0ex}}\dots$

Is this convergent?

Not convergent.

The terms in the sequence do not converge to a particular value. The value oscillates .

A general sequence is denoted by

$\phantom{\rule{2em}{0ex}}{a}_{1},\phantom{\rule{1em}{0ex}}{a}_{2},\phantom{\rule{1em}{0ex}}\dots ,\phantom{\rule{1em}{0ex}}{a}_{n},\dots$

in which ${a}_{1}$ is the first term, ${a}_{2}$ is the second term and ${a}_{n}$ is the ${n}^{th}$ term is the sequence. For example, in the harmonic sequence

$\phantom{\rule{2em}{0ex}}{a}_{1}=1,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{a}_{2}=\frac{1}{2},\dots ,{a}_{n}=\frac{1}{n}$

whilst for the alternating harmonic sequence the ${n}^{th}$ term is:

$\phantom{\rule{2em}{0ex}}{a}_{n}=\frac{{\left(-1\right)}^{n+1}}{n}$

in which ${\left(-1\right)}^{n}=+1$ if $n$ is an even number and ${\left(-1\right)}^{n}=-1$ if $n$ is an odd number.

##### Key Point 1

The sequence ${a}_{1},\phantom{\rule{1em}{0ex}}{a}_{2},\phantom{\rule{1em}{0ex}}\dots ,{a}_{n},\phantom{\rule{1em}{0ex}}\dots$ is said to be convergent if the limit of ${a}_{n}$ as $n$ increases

can be found. (Mathematically we say that $\underset{n\to \infty }{lim}{a}_{n}$ exists.)

If the sequence is not convergent it is said to be divergent .

Verify that the following sequence is convergent

$\phantom{\rule{2em}{0ex}}\frac{3}{1×2}\phantom{\rule{1em}{0ex}},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{4}{2×3}\phantom{\rule{1em}{0ex}},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{5}{3×4},\phantom{\rule{1em}{0ex}}\dots$

First find the expression for the ${n}^{th}$ term:

${a}_{n}=\frac{n+2}{n\left(n+1\right)}$

Now find the limit of ${a}_{n}$ as $n$ increases:

Hence the sequence is convergent.