In a marathon race, many runners reach the finish line. But you can also see some runners almost reaching the finish line or, in some instances, the runner barely reaching the finishing point. Similarly, in mathematics, when discussing series, you say that a series converges or diverges.
Can you also apply the same scenario you saw during the marathon? Can you check if the series converges barely or absolutely? In fact, yes, you can check if the series absolutely converges or conditionally converges.
Here, you will look at the absolute and conditional convergence of series, the difference between absolute and conditional convergence, the absolute convergence theorem, and some examples of absolute and conditional convergence.
How to define Absolute Convergence and Conditional Convergence
When it comes to series, you can determine whether they converge or diverge through different convergence tests, which you've probably seen by now. But it turns out that you can have series with both positive and negative terms. Then, how strongly do the whole series converge? This is where the concept of absolute convergence comes in.
Absolute Convergence
The absolute convergence of a series gives you the guarantee that it converges even after rearranging its terms.
If the series has both positive and negative terms and is perfectly alternating, then you can check the convergence by the alternating series test.
But what if a series is not alternating and also has negative terms? You consider the absolute value of the terms to find the convergence for this type of series.
If the series
\[\sum\limits_{n=1}^{\infty} \left | a_{n} \right | \]
converges, then the series
\[\sum\limits_{n=1}^{\infty} a_{n}\]
is called absolutely convergent, and you can say it converges absolutely.
Absolute convergence is considered to be stronger than simple convergence, as it forces you to consider the absolute values of the series. As you will see later on, absolute convergence of a series implies that the series converges as well.
Considering the series
\[\sum\limits_{n=1}^{\infty}\frac{(-1)^n}{n^3} ,\]
verify if it is absolutely convergent or not.
Solution:
To verify whether this series is absolutely convergent is to verify whether the series of terms in absolute value of the terms is convergent.
To do this, you apply the absolute value to the term \(a_n\), and identify the type of series it forms, so you know which convergence test to use.
The series of terms with absolute value of the given series is equal to a \(p-\)series with \(p=3\):
\[\sum\limits_{n=1}^{\infty}\left| \frac{(-1)^n}{n^3} \right|=\frac{1}{1^3}+\frac{1}{2^3}+\cdots+\frac{1}{n^3}=\sum\limits_{n=1}^{\infty} \frac{1}{n^3}.\]
Since \(p>1\), by the \(p-\)Series Test the given series is absolutely convergent.
For a refresher on this kind of series, see the article p-Series.
Conditional Convergence
Alternatively, what happens to the series \(\sum\limits_{n=1}^{\infty} a_{n} \) if the series of its terms with absolute value does not converge? This is where the next definition comes in.
If the series
\[\sum\limits_{n=1}^{\infty} \left | a_{n} \right | \]
diverges and the series
\[\sum\limits_{n=1}^{\infty} a_{n}\]
converges, then the series
\[\sum\limits_{n=1}^{\infty} a_{n}\]
is called conditionally convergent, and you can say it converges conditionally.
Here, the "conditionally" suggests that a series barely converges. You can check conditional convergence by testing the divergence of the series with absolute values and the convergence of the original series.
Note that both the absolute and conditional types lead to convergence. The convergence types show the series' behavior and the impact of the positive and negative terms. It does not mean that absolute convergence series converge and conditional convergence series do not converge.
Let's see this concept of conditional convergence in an example.
Show that the series
\[\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{\sqrt{n}} ,\]
is conditionally convergent.
Solution:
Here, you have two things to do:
- show that the given series is convergent;
- show that the series of terms in absolute value is divergent.
Then, you conclude that the given series is conditionally convergent.
1. The series shown is an alternating series where
\[a_n=\frac{1}{\sqrt{n}}.\]
If you check the conditions of the Alternating Series Test, you confirm that it converges. Indeed,
- \(a_n>0\) for all \(n\);
- the terms \(a_n\) decrease as \(n\) increases; and
- for the limit,\[\lim_{n \to \infty} a_n=\lim_{n \to \infty} \frac{1}{\sqrt{n}}=0.\]
Therefore, \(\sum\limits_{n=1}^{\infty}\frac{(-1)^{n+1}}{\sqrt{n}}\) converges.
For a reminder of this kind of test, see the article Alternating Series.
Now, for the convergence of\[\sum\limits_{n=1}^{\infty}\left| \frac{(-1)^{n+1}}{\sqrt{n}} \right|.\]Notice that this series of terms with absolute value is equal to a \(p-\)series with \(p=\frac{1}{2}\):\[ \sum_{n=1}^{\infty}\left| \frac{(-1)^{n+1}}{\sqrt{n}} \right|=\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}=\sum_{n=1}^{\infty} \frac{1} {n^{\frac{1}{2}}}. \]Since \(p<1\), the \(p-\)Series Test tells you that the series of terms with absolute value is divergent.
Conclusion: Since\[\sum\limits_{n=1}^{\infty}\frac{(-1)^{n+1}}{\sqrt{n}} \] converges and\[\sum\limits_{n=1}^{\infty}\left| \frac{(-1)^{n+1}}{\sqrt{n}} \right|\]diverges, the series in question is conditionally convergent (but not absolutely convergent).
Let's look at one of the important theorems about absolute convergence.
Absolute Convergence Theorem
Recall that most of the convergence tests require the terms to be positive. But with both negative and positive terms, taking absolute values and checking absolute convergence opens up more possibilities.
The absolute convergence of the series leads to the convergence of the series. In fact, that is the Absolute Convergence Theorem.
Absolute Convergence Theorem:
If the series \[\sum\limits_{n=1}^{\infty} |a_{n}|\] converges,
then the series \[\sum\limits_{n=1}^{\infty} a_{n}\] also converges.
It is important to note that the converse here is not true. That is, all convergent series may or may not be absolute.
Next, you'll see two examples, one of a direct application of the theorem, and another where the converse of the theorem doesn't hold.
Consider the series
\[\sum\limits_{n=1}^{\infty}\frac{1}{3^n-6},\]
and determine if the Absolute Convergence Theorem applies. If it does apply, what can you conclude?
Solution:
For the theorem to be applicable, you must verify the conditions of the theorem. That is, you must verify if the series of terms with absolute value converges. Looking at the absolute value you get
\[\sum\limits_{n=2}^{\infty}\left| \frac{1}{3^n-6} \right| .\]
This series is a good candidate for the Root Test. Looking at the limit,
\[\begin{align}L &=\lim_{n\to 0}\frac{\left|\dfrac{1}{3^{n+1}-6} \right|}{\left|\dfrac{1}{3^{n}-6} \right|} \\ &=\lim_{n\to 0}\frac{\left|3^{n}-6\right|}{\left|3^{n+1}-6\right|}\\ &=\lim_{n\to 0}\frac{3^{n}-6}{3^{n+1}-6}\\ &=\lim_{n\to 0}\frac{3^{n}}{3^{n+1}}\\ &=\lim_{n\to 0}\frac{1}{3}\cdot\frac{3^{n}}{3^{n}}\\ &=\frac{1}{3}.\end{align}\]
Because \(L<1\), the series
\[\sum\limits_{n=2}^{\infty}\left| \frac{1}{3^n-6} \right| \]
converges absolutely. Then, by the Absolute Convergence Theorem, you know that the series
\[\sum\limits_{n=2}^{\infty} \frac{1}{3^n-6} \]
converges.
The converse of the Absolute Convergence Theorem would be that convergence implies absolute convergence. Let's look at an example to show that this is not true.
Show that the series
\[\sum\limits_{n=2}^{\infty}\frac{(-1)^n}{\ln{n}}\]
converges conditionally.
Solution:
Let's check the convergence of the original series. The original series is an alternating series where
\[a_n=\frac{1}{\ln{n}}.\]
Checking that the conditions to use the Alternating Series Test are satisfied,
- \(\dfrac{1}{\ln{n}}>0\) for all \(n> 2\);
- the terms \(\dfrac{1}{\ln{n}}\) decrease as \(n\) increases; and
- \(\lim\limits_{n \to \infty} \frac{1}{\ln{n}}=0\).
Therefore, by the Alternating Series Test, \[\sum\limits_{n=2}^{\infty}\frac{(-1)^{n}}{\ln{n}}\] converges. Now let's look at the convergence of the series of terms with absolute value:
\[\sum\limits_{n=2}^{\infty}\left| \frac{(-1)^n}{\ln{n}} \right|=\sum\limits_{n=2}^{\infty}\frac{1}{\ln{n}}.\]
Notice that for \(n> 2\), \(\ln{n}>n\), so
\[\frac{1}{\ln{n}}\lt \frac{1}{n}.\]
You know that
\[\sum\limits_{n=2}^{\infty} \frac{1}{n} \]
is the harmonic series, so it diverges. Therefore, the series
\[\sum\limits_{n=2}^{\infty} \frac{1}{\ln{n}} \]
diverges, which means the original series does not converge absolutely. In fact the original series converges conditionally.
Therefore the convergence of a series does not imply absolute convergence of the series.
Let's take a look at a related question. What can you conclude if the series isn't absolutely convergent?
Let's go back to the previous example with the series
\[\sum\limits_{n=2}^{\infty}\frac{(-1)^n}{\ln{n}}.\]
Can you apply the Absolute Convergence Theorem to this series?
Solution:
No! In fact you saw that this series does not converge absolutely, so you can't apply the Absolute Convergence Theorem at all. This doesn't say anything about the series in question. It may converge conditionally (as it did in the previous example) or it may diverge, it just depends.
Now let's look at some more properties of series.
Difference between Absolute and Conditional Convergence
Let's take a look at a summary of the differences between absolute and conditional convergence.
Absolute Convergence Properties:
If the series \(\sum\limits_{n=1}^{\infty} \left | a_{n} \right | \) converges, then the series \(\sum\limits_{n=1}^{\infty} a_{n}\) converges absolutely;
If the series converges absolutely then both of the series \(\sum\limits_{n=1}^{\infty} a_{n}\) and \(\sum\limits_{n=1}^{\infty} \left | a_{n} \right | \) converge;
Absolute convergence is a strong condition, since absolute convergence of a series implies that the series converges.
Conditional Convergence Properties:
If the series \(\sum\limits_{n=1}^{\infty} \left | a_{n} \right | \) diverges but the series \(\sum\limits_{n=1}^{\infty} a_{n}\) converges, then the series \(\sum\limits_{n=1}^{\infty} a_{n}\) converges conditionally;
If the series converges conditionally, then \(\sum\limits_{n=1}^{\infty} \left | a_{n} \right | \) diverges;
Conditional convergence is a comparatively weaker condition since conditional convergence of a series doesn't imply that a series converges.
With that information, let's take a look at an example of absolute convergence.
Consider the following series:
\[\sum\limits_{n=1}^{\infty }\frac{\sin n}{n^2}.\]
What type of convergence (if any) does this series have?
Solution:
First let's see if the series converges absolutely by looking at
\[\sum\limits_{n=1}^{\infty}\left| \frac{\sin n}{n^2} \right|.\]
This series is a good one to use with the Direct Comparison Test.
Notice that
\[0 \le \left| \sin{n} \right| \le 1,\]
so
\[0 \le \left| \frac{\sin{n}}{n^2} \right| \le \frac{1}{n^2}.\]
The series
\[\sum\limits_{n=1}^{\infty}\frac{1}{n^2}\]
converges because it is a \(p-\)series with \(p=2\), so by the Direct Comparison Test
\[\sum\limits_{n=1}^{\infty }\left| \frac{\text{sin}\;n}{n^2} \right|\]
converges as well.
This means that the series converges absolutely, which implies that it also converges by the Absolute Convergence Theorem.
Absolute convergence is a strong convergence because just because the series of terms with absolute value converge, it makes the original series, the one without the absolute value, converge as well.
Conditional convergence is next.
Consider the series
\[\sum\limits_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}.\]
What type of convergence (if any) does this series have?
Solution:
To check it for absolute convergence, look at the series
\[\sum\limits_{n=1}^{\infty}\left| \frac{(-1)^{n+1}}{n} \right|.\]
In fact this is the harmonic series, and you know it diverges. So the original series does not converge absolutely. This means you can't use the Absolute Convergence Theorem.
Now let's check the convergence of the series. It is an alternating series where \(a_n=\dfrac{1}{n}\). The conditions to apply the Alternating Series Test are satisfied since:
- \(a_n>0\) for all \(n\);
- the terms \(a_n\) decrease as \(n\) increases; and
- \(\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \frac{1}{n}=0\).
Therefore, \(\sum\limits_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\) converges.Since the original series converges, but does not converge absolutely, it converges conditionally.
Just because the series of terms in absolute value does not converge, you
cannot conclude the original series converges – it may or may not! And if it happens to converge, then you say the series (without the terms in absolute value) converges conditionally. Conditional convergence is a weaker convergence than absolute convergence.
Next up, more examples!
Absolute Convergence Example
Let's look at another example to see how absolute convergence works.
Does the series
\[\sum\limits_{n=1}^{\infty}(-1)^{n}\frac{n^{2}+2n+5}{2^{n}}\]
converges absolutely or not?
Solution:
You can use the Alternating Series Test to show that the series converges, but that doesn't mean it converges absolutely.
To check for absolute convergence you would need to look at
\[\sum\limits_{n=1}^{\infty}\left |(-1)^{n}\frac{n^{2}+2n+5}{2^{n}} \right |=\sum\limits_{n=1}^{\infty}\frac{n^{2}+2n+5}{2^{n}}.\]
You can apply the Root Test to the series
\[\sum\limits_{n=1}^{\infty}\frac{n^{2}+2n+5}{2^{n}}\]
to see that it converges.
Recall that you state the Root Test as:
Let \(\sum\limits_{n=1}^{\infty} a_n\) be a series and define \(L\) by \[L=\lim_{n\to \infty }\left | a_n \right |^{\frac{1}{n}}=\lim_{n\to \infty }\sqrt[n]{\left | a_n \right |}.\]
Then the following hold:
1. If \(L<1\) then the series is absolutely convergent.
2. If \(L>1\) then the series diverges.
3. If \(L=1\) then the test is inconclusive.
Since the series
\[\sum\limits_{n=1}^{\infty }\left |(-1)^{n}\frac{n^{2}+2n+5}{2^{n}} \right |\] converges, the series
\[\sum\limits_{n=1}^{\infty }(-1)^{n}\frac{n^{2}+2n+5}{2^{n}}\] converges absolutely.
Let's do an example with conditional convergence next.
Conditional Convergence Example
Here you will look at the conditional convergence example.
Is the series
\[\sum\limits_{n=1}^{\infty}\frac{(-1)^{n-3}}{\sqrt{n}}\]
absolutely convergent or conditionally convergent?
Solution:
First let's check for absolute convergence by looking at
\[\sum\limits_{n=1}^{\infty}\left |\frac{(-1)^{n-3}}{\sqrt{n}} \right | = \sum\limits_{n=1}^{\infty }\frac{1}{\sqrt{n}}.\]
This series diverges by the \(p-\)Series Test. So the original series does not converge absolutely.
But will it converge conditionally? To check this, you have to find the convergence of the originally series. The original series is is an alternating series with \(a_n=\dfrac{1}{\sqrt{n}}\). The Alternating Series Test, gives you that the series
\[\sum\limits_{n=1}^{\infty}\frac{(-1)^{n-3}}{\sqrt{n}}\]
converges. Hence, the series is conditionally convergent.
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