Absolute and Conditional Convergence
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 5.5: “Alternating Series” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/5-5-alternating-series |
| Textbook used in class | Stewart, Calculus, Section 11.5: “Alternating Series” (Example 4) |
Opening Scenario
Some series converge only because positive and negative terms partially cancel each other. Rearrange those terms in a different order and the sum changes -- or the series even diverges. That fragility is captured by the distinction between absolute and conditional convergence. A series that converges absolutely is rearrangement-safe; one that converges only conditionally is not.
Quick Reference
Absolute convergence. $\sum a_n$ is absolutely convergent if $\sum |a_n|$ converges.
Conditional convergence. $\sum a_n$ is conditionally convergent if $\sum a_n$ converges but $\sum |a_n|$ diverges.
Theorem. Every absolutely convergent series is convergent: \[ \sum |a_n| \text{ converges} \implies \sum a_n \text{ converges}. \] The converse is false: $\sum a_n$ can converge without $\sum |a_n|$ converging.
Rearrangement fact (Riemann, not tested but worth knowing): a conditionally convergent series can be rearranged to converge to any real number, or to diverge. An absolutely convergent series always converges to the same sum regardless of order.
Key Concepts
1. Testing for Absolute Convergence
To test $\sum a_n$ for absolute convergence, examine $\sum |a_n|$ using any applicable convergence test (comparison, $p$-series, ratio test, etc.). If $\sum |a_n|$ converges, the original series converges absolutely.
Example 1. Is $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$ absolutely convergent? (Stewart 11.5, Example 4.)
Take absolute values: $\sum |a_n| = \sum 1/n^2$, a $p$-series with $p = 2 > 1$. It converges.
Boxed answer: The series is absolutely convergent.
2. Conditionally Convergent Example
Example 2. Show that $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$ is conditionally convergent.
The series itself converges by the Alternating Series Test ($b_n = 1/n \to 0$, decreasing).
The series of absolute values is $\sum 1/n$, the harmonic series, which diverges.
Therefore the series is conditionally convergent.
3. Hierarchy of Conclusions
| What you check | What you can conclude |
|---|---|
| $\sum \|a_n\|$ converges | $\sum a_n$ converges (absolutely) |
| $\sum \|a_n\|$ diverges | No immediate conclusion about $\sum a_n$ |
| $\sum a_n$ converges; $\sum \|a_n\|$ diverges | Conditionally convergent |
| $\sum a_n$ diverges | Neither absolute nor conditional convergence |
concluding conditional convergence from $\sum |a_n|$ diverging alone. If $\sum |a_n|$ diverges, the series might still converge (conditionally) or might diverge entirely. To conclude conditional convergence you must confirm that $\sum a_n$ itself converges. A student who sees $\sum |a_n|$ diverge and immediately writes “conditionally convergent” is skipping the essential step of checking convergence of $\sum a_n$.
Common Errors Summary
| Error | Correction |
|---|---|
| Applying absolute convergence test only to the absolute-value series, forgetting to report the original conclusion | If $\sum |a_n|$ converges, then $\sum a_n$ converges; state both |
| Concluding “conditionally convergent” when $\sum |a_n|$ diverges without checking $\sum a_n$ | First verify $\sum a_n$ converges (e.g., Alternating Series Test), then check $\sum |a_n|$ |
Common Misconceptions
if $\sum |a_n|$ diverges, then $\sum a_n$ is conditionally convergent.
This is the concept-image-conflicts-definition error about what conditional convergence requires. Conditional convergence means two things: $\sum a_n$ converges AND $\sum |a_n|$ diverges. If only $\sum |a_n|$ is known to diverge, the original series might still diverge as well. The series $\sum 1/n$ has both $\sum |1/n|$ and $\sum 1/n$ divergent, so it is neither absolutely nor conditionally convergent. Convergence of $\sum a_n$ must be established separately, typically by the Alternating Series Test, before conditional convergence can be claimed.
Leveled Practice
Problem 1. Classify $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{n^3}$ as absolutely convergent, conditionally convergent, or divergent.
Show answer
$\sum |a_n| = \sum 1/n^3$: $p$-series with $p = 3 > 1$, convergent. The original series is absolutely convergent.
Problem 2. Classify $\displaystyle\sum_{n=2}^{\infty} \frac{(-1)^n}{\ln n}$ as absolutely convergent, conditionally convergent, or divergent.
Show answer
$\sum |a_n| = \sum 1/\ln n$. Since $\ln n < n$, we have $1/\ln n > 1/n$, so by comparison with the divergent harmonic series, $\sum 1/\ln n$ diverges.
The original series $\sum (-1)^n/\ln n$: $b_n = 1/\ln n$ is decreasing and $\to 0$, so it converges by AST.
Therefore it is conditionally convergent.