Absolute and Conditional Convergence

Learning OutcomeS

Explain the meaning of absolute convergence and conditional convergence

Consider a series [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] and the related series [latex]\displaystyle\sum _{n=1}^{\infty }|{a}_{n}|[/latex]. Here we discuss possibilities for the relationship between the convergence of these two series. For example, consider the alternating harmonic series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}}{n}[/latex]. The series whose terms are the absolute value of these terms is the harmonic series, since [latex]\displaystyle\sum _{n=1}^{\infty }|\frac{{\left(-1\right)}^{n+1}}{n}|=\displaystyle\sum _{n=1}^{\infty }\frac{1}{n}[/latex]. Since the alternating harmonic series converges, but the harmonic series diverges, we say the alternating harmonic series exhibits conditional convergence.

By comparison, consider the series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}}{{n}^{2}}[/latex]. The series whose terms are the absolute values of the terms of this series is the series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}[/latex]. Since both of these series converge, we say the series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}}{{n}^{2}}[/latex] exhibits absolute convergence.

Definition

A series [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] exhibits absolute convergence if [latex]\displaystyle\sum _{n=1}^{\infty }|{a}_{n}|[/latex] converges. A series [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] exhibits conditional convergence if [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] converges but [latex]\displaystyle\sum _{n=1}^{\infty }|{a}_{n}|[/latex] diverges.

As shown by the alternating harmonic series, a series [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] may converge, but [latex]\displaystyle\sum _{n=1}^{\infty }|{a}_{n}|[/latex] may diverge. In the following theorem, however, we show that if [latex]\displaystyle\sum _{n=1}^{\infty }|{a}_{n}|[/latex] converges, then [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] converges.

theorem: Absolute Convergence Implies Convergence

If [latex]\displaystyle\sum _{n=1}^{\infty }|{a}_{n}|[/latex] converges, then [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] converges.

Proof

Suppose that [latex]\displaystyle\sum _{n=1}^{\infty }|{a}_{n}|[/latex] converges. We show this by using the fact that [latex]{a}_{n}=|{a}_{n}|[/latex] or [latex]{a}_{n}=\text{-}|{a}_{n}|[/latex] and therefore [latex]|{a}_{n}|+{a}_{n}=2|{a}_{n}|[/latex] or [latex]|{a}_{n}|+{a}_{n}=0[/latex]. Therefore, [latex]0\le |{a}_{n}|+{a}_{n}\le 2|{a}_{n}|[/latex]. Consequently, by the comparison test, since [latex]2\displaystyle\sum _{n=1}^{\infty }|{a}_{n}|[/latex] converges, the series

[latex]\displaystyle\sum _{n=1}^{\infty }\left(|{a}_{n}|+{a}_{n}\right)[/latex]

converges. By using the algebraic properties for convergent series, we conclude that

[latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}=\displaystyle\sum _{n=1}^{\infty }\left(|{a}_{n}|+{a}_{n}\right)\text{-}\displaystyle\sum _{n=1}^{\infty }|{a}_{n}|[/latex]

converges.

[latex]_\blacksquare[/latex]

Example: Absolute versus Conditional Convergence

For each of the following series, determine whether the series converges absolutely, converges conditionally, or diverges.

[latex]\displaystyle\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}}{\left(3n+1\right)}[/latex]
[latex]\displaystyle\sum _{n=1}^{\infty }\frac{\cos\left(n\right)}{{n}^{2}}[/latex]

Show Solution

try it

Determine whether the series [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{n}{\left(2{n}^{3}+1\right)}[/latex] converges absolutely, converges conditionally, or diverges.

Hint

Show Solution

Watch the following video to see the worked solution to the above Try It.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “5.5.3” here (opens in new window).

To see the difference between absolute and conditional convergence, look at what happens when we rearrange the terms of the alternating harmonic series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}}{n}[/latex]. We show that we can rearrange the terms so that the new series diverges. Certainly if we rearrange the terms of a finite sum, the sum does not change. When we work with an infinite sum, however, interesting things can happen.

Begin by adding enough of the positive terms to produce a sum that is larger than some real number [latex]M>0[/latex]. For example, let [latex]M=10[/latex], and find an integer [latex]k[/latex] such that

[latex]1+\frac{1}{3}+\frac{1}{5}+\cdots +\frac{1}{2k - 1}>10[/latex].

(We can do this because the series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{\left(2n - 1\right)}[/latex] diverges to infinity.) Then subtract [latex]\frac{1}{2}[/latex]. Then add more positive terms until the sum reaches 100. That is, find another integer [latex]j>k[/latex] such that

[latex]1+\frac{1}{3}+\cdots +\frac{1}{2k - 1}-\frac{1}{2}+\frac{1}{2k+1}+\cdots +\frac{1}{2j+1}>100[/latex].

Then subtract [latex]\frac{1}{4}[/latex]. Continuing in this way, we have found a way of rearranging the terms in the alternating harmonic series so that the sequence of partial sums for the rearranged series is unbounded and therefore diverges.

The terms in the alternating harmonic series can also be rearranged so that the new series converges to a different value. In the next example, we show how to rearrange the terms to create a new series that converges to [latex]3\text{ln}\frac{\left(2\right)}{2}[/latex]. We point out that the alternating harmonic series can be rearranged to create a series that converges to any real number [latex]r[/latex]; however, the proof of that fact is beyond the scope of this text.

In general, any series [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] that converges conditionally can be rearranged so that the new series diverges or converges to a different real number. A series that converges absolutely does not have this property. For any series [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] that converges absolutely, the value of [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] is the same for any rearrangement of the terms. This result is known as the Riemann Rearrangement Theorem, which is beyond the scope of this book.

Example: Rearranging Series

Use the fact that

[latex]1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\cdots =\text{ln}2[/latex]

to rearrange the terms in the alternating harmonic series so the sum of the rearranged series is [latex]\frac{3\text{ln}\left(2\right)}{2}[/latex].

Show Solution

Let

[latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\cdots[/latex].

Since [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}=\text{ln}\left(2\right)[/latex], by the algebraic properties of convergent series,

[latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{2}{a}_{n}=\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\cdots =\frac{1}{2}\displaystyle\sum _{n=1}^{\infty }{a}_{n}=\frac{\text{ln}2}{2}[/latex].

Now introduce the series [latex]\displaystyle\sum _{n=1}^{\infty }{b}_{n}[/latex] such that for all [latex]n\ge 1[/latex], [latex]{b}_{2n - 1}=0[/latex] and [latex]{b}_{2n}=\frac{{a}_{n}}{2}[/latex]. Then

[latex]\displaystyle\sum _{n=1}^{\infty }{b}_{n}=0+\frac{1}{2}+0-\frac{1}{4}+0+\frac{1}{6}+0-\frac{1}{8}+\cdots =\frac{\text{ln}2}{2}[/latex].

Then using the algebraic limit properties of convergent series, since [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] and [latex]\displaystyle\sum _{n=1}^{\infty }{b}_{n}[/latex] converge, the series [latex]\displaystyle\sum _{n=1}^{\infty }\left({a}_{n}+{b}_{n}\right)[/latex] converges and

[latex]\displaystyle\sum _{n=1}^{\infty }\left({a}_{n}+{b}_{n}\right)=\displaystyle\sum _{n=1}^{\infty }{a}_{n}+\displaystyle\sum _{n=1}^{\infty }{b}_{n}=\text{ln}2+\frac{\text{ln}2}{2}=\frac{3\text{ln}2}{2}[/latex].

Now adding the corresponding terms, [latex]{a}_{n}[/latex] and [latex]{b}_{n}[/latex], we see that

[latex]\begin{array}{cc}\hfill \displaystyle\sum _{n=1}^{\infty }\left({a}_{n}+{b}_{n}\right)& =\left(1+0\right)+\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(\frac{1}{3}+0\right)+\left(-\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{5}+0\right)+\left(-\frac{1}{6}+\frac{1}{6}\right)\hfill \\ \\ & +\left(\frac{1}{7}+0\right)+\left(\frac{1}{8}-\frac{1}{8}\right)+\cdots \hfill \\ & =1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\cdots .\hfill \end{array}[/latex]

We notice that the series on the right side of the equal sign is a rearrangement of the alternating harmonic series. Since [latex]\displaystyle\sum _{n=1}^{\infty }\left({a}_{n}+{b}_{n}\right)=\frac{3\text{ln}\left(2\right)}{2}[/latex], we conclude that

[latex]1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\cdots =\frac{3\text{ln}\left(2\right)}{2}[/latex].

Therefore, we have found a rearrangement of the alternating harmonic series having the desired property.

Module 5: Sequences and Series