(ST8) Absolute Convergence#

In this lesson we are going to see how to:

  • use Absolute Convergence to show that a series is convergent.

  • determine if a series \(\sum a_n\) is absolutely convergent, conditionally convergent, or divergent.

Review Video#

Absolute Convergence Test#

Absolute Convergence Test

Given a series \(\sum a_n\), we look at the related series \(\sum |a_n|\).

  • If \(\sum |a_n|\) converges, then \(\sum a_n\) converges as well.

  • If \(\sum |a_n|\) diverges, then this test is inconclusive.

Original Series
\[ \sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3+ \cdots \]

The individual terms here might be positive or negative.

Series of Absolute Values
\[ \sum_{n=1}^{\infty} |a_n| = |a_1| + |a_2| + |a_3|+ \cdots \]

The individual terms here are all positive.

Note: If all the \(a_n\) terms are positive then these two are actually the same series.

Example 1#

Determine if the following series is convergent or divergent.

\[ \sum_{n=1}^{\infty} \dfrac{(-1)^n}{n^2} \]

We want to determine the convergence of the original series:

\[ \sum a_n = \sum_{n=1}^{\infty} \dfrac{(-1)^n}{n^2} \]

To do this, we are going to look at the series of absolute value terms:

\[ \sum |a_n| = \sum_{n=1}^{\infty} \left| \dfrac{(-1)^n}{n^2}\right| \]

Since the alternating term \((-1)^n\) is inside the absolute value it effectively gets canceled off. And because \(n^2\) is always positive, we can then drop the absolute value sign to get:

\[ \sum |a_n| = \sum_{n=1}^{\infty} \dfrac{1}{n^2} \]

After simplifying, we see that \(\sum |a_n|\) is actually a convergent \(p\)-series with \(p=2\) since \(p>1\).

Since \(\sum |a_n|\) is convergent we can say that our original series \(\sum a_n\) is convergent as well by the Absolute Convergence Test.

Another way to solve this problem is to apply the Alternating Series Test directly for our original alternating series.

Definitions#

Absolutely Convergent

A series \( \sum a_n\) is said to be absolutely convergent if both \( \sum a_n\) and \(\sum |a_n|\) are convergent.

Conditionally Convergent

A series \( \sum a_n\) is said to be conditionally convergent if \( \sum a_n\) converges but \( \sum |a_n|\) diverges.

Example 2#

Determine if the following series is absolutely convergent, conditionally convergent, or divergent.

\[ \sum_{n=1}^{\infty} \dfrac{(-1)^n}{n} \]

We want to determine the convergence of the original series:

\[ \sum a_n = \sum_{n=1}^{\infty} \dfrac{(-1)^n}{n} \]

To do this, we are going to first look at the series of absolute value terms:

\[ \sum |a_n| = \sum_{n=1}^{\infty} \left| \dfrac{(-1)^n}{n}\right| \]

Since the alternating term \((-1)^n\) is inside the absolute value it effectively gets canceled off. And because \(n\) is positive for \(n\geq 1\), we can then drop the absolute value sign to get:

\[ \sum |a_n| = \sum_{n=1}^{\infty} \dfrac{1}{n} \]

After simplifying, we see that \(\sum |a_n|\) is actually a divergent \(p\)-series with \(p=1\) since \(p\leq 1\).

Since \(\sum |a_n|\) is divergent, the Absolute Convergence Test is inconclusive. This means we do not know anything about the convergence of our original series \(\sum a_n \), so we need to test this separately with a different test.

In this case, we would use the Alternating Series Test, since our original series \(\sum a_n\) has that alternating component.

\[ \sum a_n = \sum_{n=1}^{\infty} \dfrac{(-1)^n}{n} =\sum_{n=1}^{\infty} (-1)^n \cdot \dfrac{1}{n} \]

We looked at this series back in Example 1 in our Alternating Series Test lesson and found that \(\sum a_n\) was convergent.

A quick refresher on what you would need to show with the Alternating Series Test.

First identify \(b_n=\tfrac{1}{n}\), and then show:

  • Limit is 0: \(\displaystyle \lim_{n\to \infty} b_n = 0\)

  • Decreasing: \(\displaystyle b_{n+1}\leq b_n\)

Since \(\sum |a_n|\) is divergent, but \(\sum a_n\) is convergent (by the Alternating Series Test) we can say that our original series \(\sum a_n\) is conditionally convergent.

Example 3#

Determine if the following series is absolutely convergent, conditionally convergent, or divergent.

\[ \sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}}{3^n+1} \]

We want to determine the convergence of the original series:

\[ \sum a_n = \sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}}{3^n+1} \]

To do this, we are going to first look at the series of absolute value terms:

\[ \sum |a_n| = \sum_{n=1}^{\infty} \left| \dfrac{(-1)^{n-1}}{3^n+1}\right| \]

Since the alternating term \((-1)^{n-1}\) is inside the absolute value it effectively gets canceled off. And because \(3^n+1\) is positive for all \(n\geq 1\), we can drop the absolute value sign to get:

\[ \sum |a_n| = \sum_{n=1}^{\infty} \dfrac{1}{3^n+1} \]

We now see that \(\sum |a_n|\) looks comparable to a geometric series.

Since \(\sum |a_n|\) is not technically a geometric series, we need to do a little more work and use the Comparison Test (or Limit Comparison Test) to show that this is a convergent series:

\[ \sum_{n=1}^{\infty} \dfrac{1}{3^n+1} \]

Fortunately, we looked at this series back in Example 1 in our Comparison Test lesson and found that it was convergent.

A quick refresher on what you would need to show with the Comparison Test.

First identify the comparable series, in this case:

\[ \sum b_n=\sum \dfrac{1}{3^n} \]

and then show:

  • All of the terms are positive (or eventually all positive).

  • Inequality: \(\tfrac{1}{3^n+1}\leq \tfrac{1}{3^n}\)

By the Comparison Test, we found \(\sum |a_n|\) is convergent, this means \(\sum a_n\) is automatically convergent as well (by the Absolute Convergence Test).

Since both series are convergent, we can say that our original series \(\sum a_n\) is absolutely convergent.