2 General tests for convergence

The techniques we have applied to analyse the harmonic and the alternating harmonic series are ‘one-off’:- they cannot be applied to infinite series in general. However, there are many tests that can be used to determine the convergence properties of infinite series. Of the large number available we shall only consider two such tests in detail.

2.1 The alternating series test

An alternating series is a special type of series in which the sign changes from one term to the next. They have the form

$a_1-a_2+a_3-a_4+\cdots$

(in which each $a_i,i=1,2,3,\dots$ is a positive number)

Examples are:

1. $1-1+1-1+1\cdots$
2. $\frac{1}{3}-\frac{2}{4}+\frac{3}{5}-\frac{4}{6}+\cdots$
3. $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$ .

For series of this type there is a simple criterion for convergence:

Task!

Which of the following series are convergent?

1. ${\sum }_{p=1}^{\infty }{\left(-1\right)}^p\frac{\left(2p-1\right)}{\left(2p+1\right)}$
2. ${\sum }_{p=1}^{\infty }\frac{{\left(-1\right)}^{p+1}}{p^2}$

1. First, write out the series:

   Answer

   $-\frac{1}{3}+\frac{3}{5}-\frac{5}{7}+\cdots$

   Now examine the series for convergence:

   Answer

   $\frac{\left(2p-1\right)}{\left(2p+1\right)}=\frac{\left(1-\frac{1}{2p}\right)}{\left(1+\frac{1}{2p}\right)}\to 1$ as $p$ increases.

   Since the individual terms of the series do not converge to zero this is therefore a divergent series.

2. Apply the procedure used in (1) to problem (2):

   Answer

   This series $1-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+\cdots$ is an alternating series of the form $a_1-a_2+a_3-a_4+\cdots$ in which $a_p=\frac{1}{p^2}$ . The $a_p$ sequence is a decreasing sequence since $1>\frac{1}{2^2}>\frac{1}{3^2}>\dots$

   Also $\underset{p\to \infty }{lim}\frac{1}{p^2}=0$ . Hence the series is convergent by the alternating series test.