Introduction

1 Introduction

Many of the series considered in Section 16.1 were examples of finite series in that they all involved the summation of a finite number of terms. When the number of terms in the series increases without bound we refer to the sum as an infinite series . Of particular concern with infinite series is whether they are convergent or divergent. For example, the infinite series

$1 + 1 + 1 + 1 + \dots$

is clearly divergent because the sum of the first $n$ terms increases without bound as more and more terms are taken. It is less clear as to whether the harmonic and alternating harmonic series:

$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$

converge or diverge. Indeed you may be surprised to find that the first is divergent and the second is convergent. What we shall do in this Section is to consider some simple convergence tests for infinite series. Although we all have an intuitive idea as to the meaning of convergence of an infinite series we must be more precise in our approach. We need a definition for convergence which we can apply rigorously.

First, using an obvious extension of the notation we have used for a finite sum of terms, we denote the infinite series:

$a_{1} + a_{2} + a_{3} + \dots + a_{p} + \dots by the expression \sum_{p = 1}^{\infty} a_{p}$

where $a_{p}$ is an expression for the $p^{t h}$ term in the series. So, as examples:

\begin{array}{rcl} 1 + 2 + 3 + \dots & = & \sum_{p = 1}^{\infty} p since the p^{t h} term is a_{p} \equiv p \\ 1^{2} + 2^{2} + 3^{2} + \dots & = & \sum_{p = 1}^{\infty} p^{2} since the p^{t h} term is a_{p} \equiv p^{2} \\ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots & = & \sum_{p = 1}^{\infty} \frac{{(- 1)}^{p + 1}}{p} here a_{p} \equiv \frac{{(- 1)}^{p + 1}}{p} \end{array}

Consider the infinite series:

$a_{1} + a_{2} + \dots + a_{p} + \dots = \sum_{p = 1}^{\infty} a_{p}$

We consider the sequence of partial sums , $S_{1}, S_{2}, \dots$ , of this series where

\begin{array}{rcl} S_{1} & = & a_{1} \\ S_{2} & = & a_{1} + a_{2} \\ ⋮ \\ S_{n} & = & a_{1} + a_{2} + \dots + a_{n} \end{array}

That is, $S_{n}$ is the sum of the first $n$ terms of the infinite series. If the limit of the sequence $S_{1}, S_{2}, \dots, S_{n}, \dots$ can be found; that is

$lim_{n \to \infty} S_{n} = S (say)$

then we define the sum of the infinite series to be $S$ :

$S = \sum_{p = 1}^{\infty} a_{p}$

and we say “the series converges to $S$ ”. Another way of stating this is to say that

$\sum_{p = 1}^{\infty} a_{p} = lim_{n \to \infty} \sum_{p = 1}^{n} a_{p}$

Key Point 4

Convergence of Infinite Series

An infinite series $\sum_{p = 1}^{\infty} a_{p}$ is convergent if the sequence of partial sums

S_{1}, S_{2}, S_{3}, \dots, S_{k}, \dots in which S_{k} = \sum_{p = 1}^{k} a_{p} is convergent

1.1 Divergence condition for an infinite series

An almost obvious requirement that an infinite series should be convergent is that the individual terms in the series should get smaller and smaller. This leads to the following Key Point:

Key Point 5

The condition:

a_{p} \to 0 as p increases (mathematically lim_{p \to \infty} a_{p} = 0)

is a necessary condition for the convergence of the series

\sum_{p = 1}^{\infty} a_{p}

It is not possible for an infinite series to be convergent unless this condition holds.

Task!

Which of the following series cannot be convergent?

$\frac{1}{2} + \frac{2}{3} + \frac{3}{4} + \dots$
$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots$
$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$

In each case, use the condition from Key Point 5:

$a_{p} = \frac{p}{p + 1} lim_{p \to \infty} \frac{p}{p + 1} = 1$

Hence series is divergent.

$a_{p} = \frac{1}{p} lim_{p \to \infty} a_{p} = 0$

So this series may be convergent. Whether it is or not requires further testing.

$a_{p} = \frac{{(- 1)}^{p + 1}}{p} lim_{p \to \infty} a_{p} = 0$ so again this series may be convergent.

1.2 Divergence of the harmonic series

The harmonic series:

$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \dots$

has a general term $a_{n} = \frac{1}{n}$ which clearly gets smaller and smaller as $n \to \infty$ . However, surprisingly, the series is divergent. Its divergence is demonstrated by showing that the harmonic series is greater than another series which is obviously divergent. We do this by grouping the terms of the harmonic series in a particular way:

$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \dots \equiv 1 + (\frac{1}{2}) + (\frac{1}{3} + \frac{1}{4}) + (\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}) + \dots$

Now

\begin{array}{rcl} (\frac{1}{3} + \frac{1}{4}) & > & \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \\ (\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}) & > & \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{1}{2} \\ (\frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} + \frac{1}{13} + \frac{1}{14} + \frac{1}{15} + \frac{1}{16}) & > & \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} = \frac{1}{2} \end{array}

and so on. Hence the harmonic series satisfies:

$1 + (\frac{1}{2}) + (\frac{1}{3} + \frac{1}{4}) + (\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}) + \dots > 1 + (\frac{1}{2}) + (\frac{1}{2}) + (\frac{1}{2}) + \dots$

The right-hand side of this inequality is clearly divergent so the harmonic series is divergent.

1.3 Convergence of the alternating harmonic series

As with the harmonic series we shall group the terms of the alternating harmonic series, this time to display its convergence.

The alternating harmonic series is:

$S = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \dots$

This series may be re-grouped in two distinct ways.

1st re-grouping

$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} + \dots = 1 - (\frac{1}{2} - \frac{1}{3}) - (\frac{1}{4} - \frac{1}{5}) - (\frac{1}{6} - \frac{1}{7}) \dots$

each term in brackets is positive since $\frac{1}{2} > \frac{1}{3},$ $\frac{1}{4} > \frac{1}{5}$ and so on. So we easily conclude that $S < 1$ since we are subtracting only positive numbers from 1.

2nd re-grouping

$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} + \dots = (1 - \frac{1}{2}) + (\frac{1}{3} - \frac{1}{4}) + (\frac{1}{5} - \frac{1}{6}) + \dots$

Again, each term in brackets is positive since $1 > \frac{1}{2}$ , $\frac{1}{3} > \frac{1}{4},$ $\frac{1}{5} > \frac{1}{6}$ and so on.

So we can also argue that $S > \frac{1}{2}$ since we are adding only positive numbers to the value of the first term, $\frac{1}{2}$ . The conclusion that is forced upon us is that

$\frac{1}{2} < S < 1$

so the alternating series is convergent since its sum, $S$ , lies in the range $\frac{1}{2} \to 1$ . It will be shown in Section 16.5 that $S = ln 2 ≃ 0.693 .$

1 Introduction

Key Point 4

1.1 Divergence condition for an infinite series

Key Point 5

Task!

Answer

Answer

Answer

1.2 Divergence of the harmonic series

1.3 Convergence of the alternating harmonic series