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### Section 4-9 : Absolute Convergence

When we first talked about series convergence we briefly mentioned a stronger type of convergence but didn’t do anything with it because we didn’t have any tools at our disposal that we could use to work problems involving it. We now have some of those tools so it’s now time to talk about absolute convergence in detail.

First, let’s go back over the definition of absolute convergence.

#### Definition

A series \(\displaystyle \sum {{a_n}} \) is called **absolutely convergent** if \(\displaystyle \sum {\left| {{a_n}} \right|} \) is convergent. If \(\displaystyle \sum {{a_n}} \) is convergent and \(\displaystyle \sum {\left| {{a_n}} \right|} \) is divergent we call the series **conditionally convergent**.

We also have the following fact about absolute convergence.

#### Fact

If \(\displaystyle \sum {{a_n}} \) is absolutely convergent then it is also convergent.

#### Proof

First notice that \(\left| {{a_n}} \right|\) is either \({a_n}\) or it is \( - {a_n}\) depending on its sign. This means that we can then say,

\[0 \le {a_n} + \left| {{a_n}} \right| \le 2\left| {{a_n}} \right|\]Now, since we are assuming that \(\sum {\left| {{a_n}} \right|} \) is convergent then \(\sum {2\left| {{a_n}} \right|} \) is also convergent since we can just factor the 2 out of the series and 2 times a finite value will still be finite. This however allows us to use the Comparison Test to say that \(\sum \left({{a_n} + \left| {{a_n}} \right|}\right) \) is also a convergent series.

Finally, we can write,

\[\sum {{a_n}} = \sum \left({{a_n} + \left| {{a_n}} \right|}\right) - \sum {\left| {{a_n}} \right|} \]and so \(\sum {{a_n}} \) is the difference of two convergent series and so is also convergent.

This fact is one of the ways in which absolute convergence is a “stronger” type of convergence. Series that are absolutely convergent are guaranteed to be convergent. However, series that are convergent may or may not be absolutely convergent.

Let’s take a quick look at a couple of examples of absolute convergence.

- \(\displaystyle \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{n}} \)
- \(\displaystyle \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 2}}}}{{{n^2}}}} \)
- \(\displaystyle \sum\limits_{n = 1}^\infty {\frac{{\sin n}}{{{n^3}}}} \)

This is the alternating harmonic series and we saw in the last section that it is a convergent series so we don’t need to check that here. So, let’s see if it is an absolutely convergent series. To do this we’ll need to check the convergence of.

\[\sum\limits_{n = 1}^\infty {\left| {\frac{{{{\left( { - 1} \right)}^n}}}{n}} \right|} = \sum\limits_{n = 1}^\infty {\frac{1}{n}} \]This is the harmonic series and we know from the integral test section that it is divergent.

Therefore, this series is not absolutely convergent. It is however conditionally convergent since the series itself does converge.

b \(\displaystyle \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 2}}}}{{{n^2}}}} \) Show Solution

In this case let’s just check absolute convergence first since if it’s absolutely convergent we won’t need to bother checking convergence as we will get that for free.

\[\sum\limits_{n = 1}^\infty {\left| {\frac{{{{\left( { - 1} \right)}^{n + 2}}}}{{{n^2}}}} \right|} = \sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} \]This series is convergent by the \(p\)-series test and so the series is absolute convergent. Note that this does say as well that it’s a convergent series.

c \(\displaystyle \sum\limits_{n = 1}^\infty {\frac{{\sin n}}{{{n^3}}}} \) Show Solution

In this part we need to be a little careful. First, this is NOT an alternating series and so we can’t use any tools from that section.

What we’ll do here is check for absolute convergence first again since that will also give convergence. This means that we need to check the convergence of the following series.

\[\sum\limits_{n = 1}^\infty {\left| {\frac{{\sin n}}{{{n^3}}}} \right|} = \sum\limits_{n = 1}^\infty {\frac{{\left| {\sin n} \right|}}{{{n^3}}}} \]To do this we’ll need to note that

\[ - 1 \le \sin n \le 1\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}\left| {\sin n} \right| \le 1\]and so we have,

\[\frac{{\left| {\sin n} \right|}}{{{n^3}}} \le \frac{1}{{{n^3}}}\]Now we know that

\[\sum\limits_{n = 1}^\infty {\frac{1}{{{n^3}}}} \]converges by the \(p\)-series test and so by the Comparison Test we also know that

\[\sum\limits_{n = 1}^\infty {\frac{{\left| {\sin n} \right|}}{{{n^3}}}} \]converges.

Therefore, the original series is absolutely convergent (and hence convergent).

Let’s close this section off by recapping a topic we saw earlier. When we first discussed the convergence of series in detail we noted that we can’t think of series as an infinite sum because some series can have different sums if we rearrange their terms. In fact, we gave two rearrangements of an Alternating Harmonic series that gave two different values. We closed that section off with the following fact,

#### Facts

Given the series \(\displaystyle \sum {{a_n}} \),

- If \(\displaystyle \sum {{a_n}} \) is absolutely convergent and its value is \(s\) then any rearrangement of \(\sum {{a_n}} \) will also have a value of \(s\).
- If \(\displaystyle \sum {{a_n}} \) is conditionally convergent and \(r\) is any real number then there is a rearrangement of \(\sum {{a_n}} \) whose value will be \(r\).

Now that we’ve got the tools under our belt to determine absolute and conditional convergence we can make a few more comments about this.

First, as we showed above in Example 1a an Alternating Harmonic is conditionally convergent and so no matter what value we chose there is some rearrangement of terms that will give that value. Note as well that this fact does not tell us what that rearrangement must be only that it does exist.

Next, we showed in Example 1b that,

\[\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 2}}}}{{{n^2}}}} \]is absolutely convergent and so no matter how we rearrange the terms of this series we’ll always get the same value. In fact, it can be shown that the value of this series is,

\[\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 2}}}}{{{n^2}}}} = - \frac{{{\pi ^2}}}{{12}}\]