Title: Default for 16.2 Infinite Series
Course: HELM Workbooks 2023-2024

Exercise

Use the ratio test alone to examine the convergence of the series:

1. \(\displaystyle \sum_{p=1}^\infty \frac{e^[value]}{(2p+1)^{p+1}}\). The ratio test is .
2. \(\displaystyle \sum_{p=1}^\infty \frac{p^[value]}{p!}\). The ratio test is .
3. \(\displaystyle \sum_{p=1}^\infty \frac{1}{\sqrt{p}}\). The ratio test is .
4. \(\displaystyle \sum_{p=1}^\infty \frac{1}{([value])^p}\). The ratio test is .
5. \(\displaystyle \sum_{p=1}^\infty \frac{(-1)^{p+1}}{[value]^p}\). The ratio test is .
Feedback: 1. The general term of this series is \(\displaystyle a_p = \frac{e^[value]}{(2p + 1)^{p+1}}\). Then \[\begin{aligned} \lambda = \lim_{p \to \infty} \frac{a_{p+1}}{a_p} &= \lim_{p \to \infty} \frac{e^[value]}{(2(p+1) + 1)^{p+2}} \cdot \frac{(2p + 1)^{p+1}}{e^[value]} \\ &= \lim_{p \to \infty} \frac{(2p + 1)^{p+1}}{(2(p+1) + 1)^{p+2}} \\ &= \lim_{p \to \infty} \left(\frac{2p + 1}{2p + 3}\right)^{p+1} \cdot \frac{1}{2p + 3} \\ &\leq 1 \cdot \lim_{p \to \infty} \frac{1}{2p + 3} \\ &= 1 \cdot 0 = 0. \end{aligned}\] But the limit is clearly nonnegative, so we conclude that \(\lambda = 0\). Since \(\lambda < 1\), the series converves. 2. The general term of this series is \(\displaystyle a_p = \frac{p^[value]}{p!}\). Then \[\begin{aligned} \lambda = \lim_{p \to \infty} \frac{a_{p+1}}{a_p} &= \lim_{p \to \infty} \frac{(p+1)^[value]}{(p+1)!} \cdot \frac{p!}{p^[value]} \\ &= \lim_{p \to \infty} \left(\frac{p+1}{p}\right)^[value] \cdot \frac{1}{p+1} \\ &= 1 \cdot 0 = 0. \end{aligned}\] Since \(\lambda < 1\), the series converges. 3. The general term of this series is \(\displaystyle a_p = \frac{1}{\sqrt{p}}\). Then \[\begin{aligned} \lambda = \lim_{p \to \infty} \frac{a_{p+1}}{a_p} &= \lim_{p \to \infty} \frac{1}{\sqrt{p+1}} \cdot \frac{\sqrt{p}}{1} \\ &= \lim_{p \to \infty} \frac{\sqrt{p}}{\sqrt{p+1}} \\ &= \lim_{p \to \infty} \sqrt{\frac{p}{p+1}} \\ &= 1. \end{aligned}\] Since \(\lambda = 1\), the ratio test is inconclusive. Nevertheless, note that \(\displaystyle \frac{1}{\sqrt{p}} > \frac{1}{p}\) for all \(p\), so our series diverges by comparison with the harmonic series. 4. The general term of this series is \(\displaystyle a_p = \frac{1}{([value])^p}\). Then \[\begin{aligned} \lambda = \lim_{p \to \infty} \frac{a_{p+1}}{a_p} &= \lim_{p \to \infty} \frac{1}{([value])^{p+1}} \cdot \frac{([value])^p}{1} \\ &= \lim_{p \to \infty} \frac{([value])^{p}}{([value])^{p+1}} \\ &= \lim_{p \to \infty} \frac{1}{[value]} \\ &= \frac{1}{[value]} \approx [value]. \end{aligned}\] Since \(\lambda > 1\), the series diverges. 5. The general term of this series is \(\displaystyle a_p = \frac{(-1)^{p+1}}{[value]^p}\). Note that this is not always positive, so we cannot apply the ratio test. It is thus inconclusive. Nevertheless, this series is absolutely convergent, and thus convergent.

Exercise

For what values of \(x\) are the following series absolutely convergent?

1. \(\displaystyle{\sum_{p=1}^\infty \frac{(-1)^p x^p}{p}}\) is convergent for all \(|x|\)  .
2. \(\displaystyle{\sum_{p=1}^\infty \frac{(-1)^p x^p}{p!}}\) is convergent for all \(|x|\) .

Hint: if the series is absolutely convergent for all values of \(x\), select = and then enter R or all (to signify the real numbers).
Feedback: 1. The general term of the related series of positive terms is \(\displaystyle a_p = \frac{|x|^p}{p}\). For this series, using the ratio test we find \[\begin{aligned} \lambda = \lim_{p \to \infty} \frac{a_{p+1}}{a_p} &= \lim_{p \to \infty} \frac{|x|^{p+1}}{p+1} \cdot \frac{p}{|x|^p} \\ &= \lim_{p \to \infty} \frac{p}{p+1} \cdot |x| \\ &= |x|. \end{aligned}\] Hence, the original series is absolutely convergent for all \(|x| < 1\). Note that if \(|x| = 1\), then the series of positive terms is the harmonic series, which diverges. Therefore, our series is absolutely convergent if and only if \(|x| < 1\). 2. The general term of the related series of positive terms is \(\displaystyle a_p = \frac{|x|^p}{p!}\). For this series, using the ratio test we find \[\begin{aligned} \lambda = \lim_{p \to \infty} \frac{a_{p+1}}{a_p} &= \lim_{p \to \infty} \frac{|x|^{p+1}}{(p+1)!} \cdot \frac{p!}{|x|^p} \\ &= \lim_{p \to \infty} \frac{p!}{(p+1)!} \cdot |x| \\ &= \lim_{p \to \infty} \frac{1}{p+1} \cdot |x| \\ &= 0 \cdot |x| = 0. \end{aligned}\] We have found that \(\lambda = 0 < 1\) irrespective of the value of \(|x|\). Hence, our series is absolutely convergent for all \(|x|\) in \(\mathbb{R}\).

Infinite series

Many of the series considered in the previous Section 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: \[ 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots \qquad \qquad 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots \] 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: \[ a_1 + a_2 + a_3 + \dots + a_p + \dots \quad\mathrm{by \: the \: expression } \; \sum_{p=1}^{\infty} a_p\] where \(a_p\) is an expression for the \(p^{th}\) term in the series. So, as examples:

\[1+2+3+\dots = \sum_{p=1}^{\infty} p \quad \mathrm{since\: the \:} p^{th} \: \mathrm{ term \: is } \: a_p \equiv p \]

\[1^2+2^2+3^2+\dots = \sum_{p=1}^{\infty} p^2 \quad \mathrm{since \: the \:} p^{th} \: \mathrm{ term \: is } \: a_p \equiv p^2 \]

\[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots = \sum_{p=1}^{\infty} \dfrac{(-1)^{p+1}}{p} \quad \mathrm{here \: } a_p \equiv \dfrac{(-1)^{p+1}}{p} \]

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 \[ \begin{aligned} S_1 & = a_1 \\ S_2 & = a_1 + a_2 \\ \vdots & \\ S_n & = a_1 + a_2 + \dots + a_n \end{aligned}\]

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

\[ \lim_{n \to \infty} S_n = S \quad \mathrm{(say)}\] 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 \[\sum_{p=1}^{\infty} a_p =\lim_{n \to \infty} \sum_{p=1}^{\infty} a_p\]

Key Point

An infinite series \(\sum_{p=1}^{\infty} a_p\) is convergent if the sequence of partial sums \[S_1, S_2, S_3,\dots,S_k,\dots \qquad \mathrm{in \: which} \quad S_k = \sum_{k=1}^{\infty} \: a_k \:\mathrm{is \: convergent.} \]

The divergence test

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

The condition: \[ a_p \to 0 \qquad \mathrm{as} \; p \; \mathrm{increases} \quad \left(\text{mathematically:} \lim_{p \to \infty} a_p = 0\right)\] is a necessary condition for the convergence of the series \(\displaystyle{ \sum_{p=1}^{\infty} a_p}\) .

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

We need to be careful with the logic here. If a series converges then its terms tend to zero. If we observe that the terms of a series tend to zero then we cannot make any conclusion about the series. It might converge, or it might diverge (see the harmonic series).

If we reverse the logic, called taking the contrapositive, we have the following key point.

Key Point

Given the series \[\sum_{p=1}^\infty a_p\] we can negate the condition above \[ a_p \nrightarrow 0 \qquad \mathrm{as} \; p \; \mathrm{increases} \quad \left(\text{mathematically:} \lim_{p \to \infty} a_p \neq 0\text{ or }\lim_{p \to \infty} a_p\text{ does not exist.}\right) \] When this condition is satisfied we can conclusively say that the series is divergent. This is called the divergence test.

Remember that not all divergent series satisify the divergence test.

Exercise

Consider the following series. We will apply the key point above to see if we can determine if any of the series diverge.

1. \( \displaystyle{\sum_{p=1}^\infty a_p= \dfrac{1}{2} + \dfrac{2}{3} + \dfrac{3}{4} + \dots}\)
\(a_p = \)
\(\displaystyle{\lim_{p \to \infty} a_p =} \)

2. \(\displaystyle{\sum_{p=1}^\infty b_p= 1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dots}\)
\(b_p = \)
\(\displaystyle{\lim_{p \to \infty} b_p= }\)

3. \(\displaystyle{\sum_{p=1}^\infty c_p=1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dots} \)
\(c_p = \)
\(\displaystyle{\lim_{p \to \infty} c_p= }\)

4. Which of the series are divergent by the divergence test?
Feedback: 1. \(a_p = \dfrac{p}{p+1}\) \( \lim_{p \to \infty}\dfrac{p}{p+1} = 1 \neq 0\), therefore the series is divergent. 2. \(b_p = \dfrac{1}{p}\) \(\lim_{p \to \infty}\dfrac{1}{p} = 0\), so this series may be convergent. Whether it is or not requires further testing. 3. \(c_p = \dfrac{(-1)^{p+1}}{p}\) \(\lim_{p \to \infty}\dfrac{(-1)^{p+1}}{p} = 0 \), so again this series may be convergent. Whether it is or not requires further testing. 4. Only \(\displaystyle{\sum_{p=1}^\infty a_p}\) is divergent by the divergence test. Note that exactly one of the other series is also divergent!

The harmonic series:

\[ 1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + \dots\]

has a general term \(a_n=\dfrac{1}{n}\) 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:

\[ 1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + \dots \equiv 1 + \left(\dfrac{1}{2}\right) +  \left(\dfrac{1}{3} + \dfrac{1}{4}\right) +  \left(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\right) + \dots\]

Now \[ \begin{aligned} \left(\dfrac{1}{3} + \dfrac{1}{4}\right) & > \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2} \\  \left(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\right) & > \dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8} = \dfrac{1}{2} \\ \left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12} + \dfrac{1}{13} + \dfrac{1}{14}+ \dfrac{1}{15}+ \dfrac{1}{16}\right) & > \left(\dfrac{1}{16}+\dfrac{1}{16}+\dfrac{1}{16}+\dfrac{1}{16} + \dfrac{1}{16} + \dfrac{1}{16}+ \dfrac{1}{16}+ \dfrac{1}{16}\right) = \dfrac{1}{2} \end{aligned}\]

and so on. Hence the harmonic series satisfies:

\[\begin{aligned}1+\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + \dots & \equiv 1 + \left(\dfrac{1}{2}\right) +  \left(\dfrac{1}{3} + \dfrac{1}{4}\right) +  \left(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\right)+ \dots \\ & >1 + \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} + \dots \end{aligned}\]

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

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:

\[ S = 1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{5} - \dots\]

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

1st re-grouping

\[ S = 1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{5} + \dots = 1 - \left(\dfrac{1}{2} - \dfrac{1}{3} \right) - \left( \dfrac{1}{4} - \dfrac{1}{5} \right) - \left(  \dfrac{1}{6} - \dfrac{1}{7}\right) - \dots\]

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

2nd re-grouping

\[ S = 1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{5} + \dots = \left(1 - \dfrac{1}{2}\right) + \left(\dfrac{1}{3} - \dfrac{1}{4}\right) + \left(\dfrac{1}{5} - \dfrac{1}{6}\right) + \dots \]

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

So we can also argue that \(S > \dfrac{1}{2}\) since we are adding only positive numbers to the value of the first term, \(\dfrac{1}{2}\) .

The conclusion that is forced upon us is that \[\dfrac{1}{2}<S<1\] so that the alternating series is convergent since its sum, \(S\), lies in the range \(\dfrac{1}{2} \to 1\). It will be shown in a later Section that \(S = \ln{2} = 0.693\).

The alternating series test and ratio test

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 many infinite series. Of the large number available we shall only consider two such tests in detail, namely the ratio test and alternating series test.

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

\[a_1 - a_2 + a_3 - a_4 + \dots\]

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

Examples are:

1. \(1-1+1-1+\dots\)
2. \(\dfrac{1}{3} - \dfrac{2}{4} + \dfrac{3}{5} - \dfrac{4}{6} + \dots\)
3. \(1-\dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4}+\dots\)

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

Key Point:

The Alternating Series Test

The alternating series \[a_1-a_2+a_3-a_4+\dots\] (in which \(a_i, \; i = 1,2,3,\dots\) are positive numbers) is convergent if both

- the terms continually decrease: \[a_1>a_2>a_3>\dots\]
- and the terms decrease to zero: \[a_p \to 0 \: \mathrm{as \;} p \mathrm{\: increases} \qquad \left(\mathrm{mathematically \:} \lim_{p \to \infty} a_p=0\right)\]

Note that if the series fails the alternating series test then it may still be convergent, you simply are not equipped to determine the outcome.

Exercise

Which of the following series are convergent?

1. The series \(\displaystyle{\sum_{p=1}^{\infty}(-1)^p \dfrac{2p-1}{2p+1}}\) is
2. The series \(\displaystyle{\sum_{p=1}^{\infty}\dfrac{(-1)^{p+1}}{p^2}} \) is
Feedback: 1. First, write out the series \[ -\dfrac{1}{3} + \dfrac{3}{5} -\dfrac{5}{7} + \dots\] Now examine the series for convergence: \[ \dfrac{2p-1}{2p+1} = \dfrac{\left(1-\frac{1}{2p}\right)}{\left(1+\frac{1}{2p}\right)} \to 1 \qquad \mathrm{as \:} p \: \mathrm{increases.} \] Since the individual terms of the series do not converge to zero this is therefore a divergent series. 2. This series \[ 1 - \dfrac{1}{2^2} + \dfrac{1}{3^2} - \dfrac{1}{4^2}+ \dots \] is an alternating series of the form \(a_1-a_2+a_3-a_4+\dots\) in which \(a_p=\dfrac{1}{p^2}\). The \(a_p\) sequence is a decreasing sequence since \(1 > \dfrac{1}{2^2} > \dfrac{1}{3^2} > \dots\) Also \(\lim_{p \to \infty} \dfrac{1}{p^2} = 0\). Hence the series is convergent by the alternating series test.

The ratio test (also known as d'Alembert's ratio test)

The below test, which is one of the most useful and widely used convergence tests, applies only to series of positive terms.

Key Point

The Ratio Test

Let \(\displaystyle{\sum_{p=1}^{\infty} a_p} \) be a series of positive terms such that, as \(p\) increases, the limit of \(\dfrac{a_{p+1}}{a_p}\) equals a number \(\lambda\). That is \(\displaystyle{\lim_{p \to \infty} \dfrac{a_{p+1}}{a_p} = \lambda}\).

It can be shown that:

- if \(\lambda >1 \quad\) then  \(\displaystyle{\sum_{p=1}^{\infty} a_p }\) diverges.
- if \(\lambda<1 \quad \) then  \(\displaystyle{\sum_{p=1}^{\infty} a_p }\) converges.
- if \(\lambda = 1 \quad\) then  \(\displaystyle{\sum_{p=1}^{\infty} a_p} \) may converge or diverge.
That is, the test is inconclusive in this case.

Example

Use the ratio test to examine the convergence of the series

1. \(\displaystyle{\sum_{p=1}^\infty a_p=} 1 + \dfrac{1}{2!} + \dfrac{1}{3!} + \dfrac{1}{4!}+ \dots\)
2. \( \displaystyle{\sum_{p=1}^\infty b_p=}1 + x^2 + x^3  + \dots\)

Solution

1. The general term in this series is \(\dfrac{1}{p!}\) i.e. \[ 1+ \dfrac{1}{2!} + \dfrac{1}{3!} + \dots = \sum_{p=1}^{\infty} \dfrac{1}{p!} \] So \[ a_p = \dfrac{1}{p^2} \quad \therefore \quad a_{p+1} = \dfrac{1}{(p+1)!}\] and the ratio \[\dfrac{a_{p+1}}{a_p} = \dfrac{p!}{(p+1)!}= \dfrac{p(p-1) \dots (3)(2)(1)}{(p+1)p(p_1) \dots (3)(2)(1)}= \dfrac{1}{p+1}\] \[\therefore \quad \lim_{p \to \infty} \dfrac{a_{p+1}}{a_p} = \lim_{p \to \infty} \dfrac{1}{p+1} = 0\] Since \(0<1\) the series is convergent. In fact, it can be shown that \[1 + \dfrac{1}{2!} + \dfrac{1}{3!} + \dots = e - 1 \approx 1.718\]
2. Here we must assume that \(x > 0\) since we can only apply the ratio test to a series of positive terms. Now \[ 1 + x + x^2 + x^3 + \dots = \sum_{p=1}^{\infty} x^{p-1}\] so that \[b_p=x^{p-1}, \quad a_{p+1} = x^p\] and \[ \lim_{p \to \infty} \dfrac{b_{p+1}}{a_p} = \lim_{p \to \infty} \dfrac{x^p}{x^{p+1}} = \lim_{p \to \infty} x = x\] Thus, using the ratio test we deduce that (if \(x\) is a positive number) this series will only converge if \(x < 1\). We will see in a later Section that \[1+x+x^2+x^3+\dots = \dfrac{1}{1-x} \qquad \text{provided} \; 0<x<1\]

Exercise

Use the ratio test to examine the convergence of the series: \[\sum_{p=1}^\infty a_p =\dfrac{1}{\ln [value]}+\dfrac{8}{(\ln [value])^2}+ \dfrac{27}{(\ln [value])^3} + \dots\]

Note: in the input boxes below, type log(x) to represent \( \ln x \) (the natural logarithm of \(x\)).

First, find the general term of the series:

\(a_p = \)

Now find \(a_{p+1}\):

\(a_{p+1} = \)

Finally, obtain \( \lim_{p \to \infty} \dfrac{a_{p+1}}{a_p}\):

\(\dfrac{a_{p+1}}{a_p}= \) \(\qquad \therefore \quad \lim_{p \to \infty} \dfrac{a_{p+1}}{a_p}=\)

Hence the ratio test indicates that this series is
Feedback: \[ \dfrac{1}{\ln [value]} + \dfrac{8}{(\ln [value])^2} + \dots = \sum_{p=1}^{\infty} \dfrac{p^3}{(\ln [value])^p}\] So \[a_p = \dfrac{p^3}{(\ln [value])^p}\] and \[a_{p+1}=\dfrac{(p+1)^3}{(\ln [value])^{p+1}}\] Therefore, \[ \dfrac{a_{p+1}}{a_p} = \left(\dfrac{p+1}{p}\right)^3 \dfrac{1}{\ln [value]}\] Now \[\left(\dfrac{p+1}{p}\right)^3 = \left(1 + \dfrac{1}{p}\right)^3 \to 1 \quad \mathrm{as \:} p \:\mathrm{increases}\] \[ \therefore \lim_{p \to \infty} \dfrac{a_{p+1}}{a_p} = \dfrac{1}{\ln [value]} < 1\] Hence this is a convergent series.

Note that in all of these examples and exercises we have decided upon the convergence or divergence of various series; we have not been able to use the tests to discover what actual number the convergent series converges to.

Conditional vs absolute convergence

The ratio test applies to series of positive terms. Indeed this is true of many related tests for convergence. However, as we have seen, not all series are series of positive terms. To apply the ratio test such series must first be converted into series of positive terms. This is easily done.

Consider two series \[\sum_{p=1}^{\infty} a_p \qquad \mathrm{and} \qquad  \sum_{p=1}^{\infty} |a_p|.\] The latter series, obviously directly related to the first, is a series of positive terms. Note that is by the use of absolute values that the we have the name absolute convergence, as we'll soon define.

Using imprecise language, it is harder for the second series to converge than it is for the first, since, in the first, some of the terms may be negative and cancel out part of the contribution from the positive terms. No such cancellations can take place in the second series since they are all positive terms. Thus it is plausible that if \(\displaystyle{\sum_{p=1}^{\infty} |a_p|}\) converges then so does \(\displaystyle{\sum_{p=1}^{\infty} a_p}\).

This leads to the following definition.

Key Point

A convergent series \(\displaystyle{\sum_{p=1}^{\infty} a_p}\) is said to be conditionally convergent if \(\displaystyle{\sum_{p=1}^{\infty} |a_p|}\) is divergent.

A convergent series \(\displaystyle{\sum_{p=1}^{\infty} a_p}\) is said to be absolutely convergent if \(\displaystyle{\sum_{p=1}^{\infty} |a_p|}\) is convergent.

For example, the alternating harmonic series: \[ \sum_{p=1}^{\infty} \dfrac{(-1)^{p+1}}{p} = 1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dots \] is conditionally convergent since the series of positive terms (the harmonic series): \[ \sum_{p=1}^{\infty} \left|\dfrac{(-1)^{p+1}}{p}\right| \equiv \sum_{p=1}^{\infty} \dfrac{1}{p} = 1 +\dfrac{1}{2} + \dfrac{1}{3} + \dots\] is divergent.

For any series we have three possibilities; absolutely convergent, conditionally convergent, or divergent. Every series will fall into exactly one of these categories.

Exercise

Examine the convergence of the series \[\sum_{p=1}^\infty a_p= -\dfrac{1}{2!} + \dfrac{1}{4!} - \dfrac{1}{6!} + \dots \]

First, find the general term of the series:

\(a_p=\)

Write down an expression for the general term of the related series of positive terms:

\(|a_p|=\)

Now use the ratio test to examine the convergence of this series:

\(\dfrac{|a_{p+1}|}{|a_p|}=\) , therefore \(\lim_{p \to \infty} \dfrac{|a_{p+1}|}{|a_p|}=\)

Hence the series is
Feedback: \[-\dfrac{1}{2!} + \dfrac{1}{4!} - \dfrac{1}{6!} + \dots = \sum_{p=1}^{\infty} \dfrac{(-1)^p}{(2p)!}\] So \(a_p = \dfrac{(-1)^p}{(2p)!}\) is the general term of the series, and therefore \(|a_p| = \dfrac{1}{(2p)!}\). \[ \begin{aligned} \dfrac{|a_{p+1}|}{|a_p|} & = \dfrac{(2p)!}{(2(p+1)!)} \\ & = \dfrac{2p(2p-1)\dots(2)(1)}{(2p+2)(2p+1)(2p)(2p-1)\dots(2)(1) } \\ & = \dfrac{1}{(2p+2)(2p+1)} \to 0 \quad \mathrm{as} \: p \: \mathrm{increases}\end{aligned}\] So the series of positive terms is convergent by the ratio test. Hence \(\sum_{p=1}^{\infty} \dfrac{(-1)^p}{(2p)!}\) is absolutely convergent.

Exercise

Which of the following alternating series are convergent?

1. \(\displaystyle \sum_{p=1}^{\infty} \frac{(-1)^p \ln ([value])}{p}\) is .
2. \(\displaystyle \sum_{p=1}^{\infty} \frac{(-1)^{p+1}}{p^2 +1}\) is .
3. \(\displaystyle \sum_{p=1}^{\infty} \frac{p \sin \left((2p+1)\dfrac{\pi}{2}\right)}{p+[value]}\) is .
Feedback: 1. We write the series out: \[-\frac{\ln [value]}{1} + \frac{\ln [value]}{2} - \frac{\ln [value]}{3} + \cdots\] Notice that the series is alternating, and its terms are continuously decreasing to zero. The alternating series test then implies that the series converges. 2. We write the series out: \[\frac{1}{1^2 + 1} - \frac{1}{2^2 + 1} + \frac{1}{3^2 + 1} - \cdots\] Notice that the series is alternating, and its terms are continuously decreasing to zero. The alternating series test then implies that the series converges. 3. First notice that we can rewrite the argument to the \(\sin\) function in the general term as \(p\pi + \pi/2\). We see then that when \(p\) is odd, the sine term becomes \(-1\), while when \(p\) is even, it becomes \(1\). We can thus write the series out: \[-\frac{1}{1 + [value]} + \frac{2}{2 + [value]} - \frac{3}{3 + [value]} \cdots\] Note that the general term of this series does not converge to zero. As this is a necessary condition for convergence, we conclude that the series diverges.