Absolute and Conditional Convergence

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.

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:

1. show that the given series is convergent;
2. show that the series of terms in absolute value is divergent.

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,\[\lim_{n \to \infty} a_n=\lim_{n \to \infty} \frac{1}{\sqrt{n}}=0.\]

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.

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
• \(\lim\limits_{n \to \infty} \frac{1}{\ln{n}}=0\).

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 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.

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.

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
• \(\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \frac{1}{n}=0\).

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.

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.

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.