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Section 8.8 Absolute Convergence (SQ8)

Subsection 8.8.1 Activities

Activity 8.8.1.

Recall the series \(\displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{n}\) from Activity 8.7.5.
(a)
Does the series \(\displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{n}\) converge or diverge?
(b)
Does the series \(\displaystyle \sum_{n=1}^\infty \left|\frac{(-1)^n}{n}\right|\) converge or diverge?

Activity 8.8.2.

Consider the series \(\displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{n^2}\text{.}\)
(a)
Does the series \(\displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{n^2}\) converge or diverge?
(b)
Does the series \(\displaystyle \sum_{n=1}^\infty \left|\frac{(-1)^n}{n^2}\right|\) converge or diverge?

Definition 8.8.3.

Given a series
\begin{equation*} \sum a_n \end{equation*}
we say that \(\displaystyle \sum a_n\) is absolutely convergent if \(\displaystyle \sum |a_n|\) converges.

Activity 8.8.4.

Consider the series: \(\displaystyle \sum_{n=1}^\infty \frac{(-1)^nn!}{(2n)!}\text{.}\)
(a)
Does the series \(\displaystyle \sum_{n=1}^\infty \frac{(-1)^nn!}{(2n)!}\) converge or diverge? (Recall Fact 8.7.6.)
(b)
Compute \(|a_n|\text{.}\)
(c)
Does the series \(\displaystyle \sum_{n=1}^\infty \frac{(-1)^nn!}{(2n)!}\) converge absolutely?

Activity 8.8.6.

Consider the series: \(\displaystyle \sum_{n=1}^\infty -n\text{.}\)
(a)
Does the series \(\displaystyle \sum_{n=1}^\infty -n\) converge or diverge?
(b)
Compute \(|a_n|\text{.}\)
(c)
Does the series \(\displaystyle \sum_{n=1}^\infty -n\) converge absolutely?

Activity 8.8.7.

For each of the following series, determine if the series is convergent, and if the series is absolutely convergent.
(a)
\(\displaystyle \sum_{n=1}^\infty \frac{n^2(-1)^n}{n^3+1}\)
(b)
\(\displaystyle \sum_{n=1}^\infty \frac{1}{n^2}\)
(c)
\(\displaystyle \sum_{n=1}^\infty (-1)^n \left(\frac{2}{3}\right)^n\)

Activity 8.8.8.

If you know a series \(\displaystyle \sum a_n\) is absolutely convergent, what can you conclude about whether or not \(\displaystyle \sum a_n\) is convergent?
  1. We cannot determine if \(\displaystyle \sum a_n\) is convergent.
  2. \(\displaystyle \sum a_n\) is convergent since it “grows slower” than \(\displaystyle \sum |a_n|\) (and falls slower than \(\displaystyle \sum -|a_n|\)).

Subsection 8.8.2 Videos

Figure 184. Video: Determine if a series converges absolutely or conditionally