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Section 8.8 Absolute Convergence (SQ8)

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 8.8.1 Videos

Figure 158. Video: Determine if a series converges absolutely or conditionally