Classify convergence correctly
One family, three verdicts
The series is \(1-\dfrac{1}{2^p}+\dfrac{1}{3^p}-\cdots\), and the slider sets the exponent \(p\). The bars show the terms, green for positive and red for negative. The size \(|a_n|\) of a term is its magnitude. The orange dots show the running total of the sizes, and the teal dots show the running total with signs.
At p = 2, both the orange and the teal dots level off. The series and its magnitudes both converge, so the series is absolutely convergent.
Move p to 0.5. The teal dots still approach a value, but the orange dots rise without limit and leave the graph. The series converges and its magnitudes diverge, so it is conditionally convergent.
Move p to 0. Every term has size 1, so the teal dots jump between 1 and 0 and do not approach a value. The series diverges.
The magnitudes \(\frac{1}{n^p}\) form a p-series, so the series is absolutely convergent only for \(p>1\).
Two questions, three cases
Two questions sort every series. Do the orange dots level off? Do the teal dots approach a value? Each button shows one case.
If the orange dots level off, the series is absolutely convergent.
If the orange dots rise without limit and the teal dots approach a value, the series is conditionally convergent.
If the teal dots do not approach a value, the series is divergent.
Absolute convergence forces convergence, so orange dots that level off never appear with teal dots that fail to approach a value.
Notation
A series is absolutely convergent when the series of magnitudes converges, so the orange dots level off.
\[\cy{\sum|a_n|}\ \text{converges}\]A series is conditionally convergent when it converges and the series of magnitudes diverges. The teal dots approach a value, but the orange dots rise without limit.
\[\begin{aligned}\cb{\sum a_n}\ &\text{converges},\\ \cy{\sum|a_n|}\ &\text{diverges}\end{aligned}\]A series is divergent when the series itself diverges. Then the series of magnitudes diverges too.
\[\cb{\sum a_n}\ \text{diverges}\]To classify a series, first test the magnitudes. If they converge, stop, because the series is absolutely convergent.
If the magnitudes diverge, test the original series. If it converges, the series is conditionally convergent. Otherwise it is divergent.
The names matter when the terms are added in a different order. An absolutely convergent series keeps its sum, but a conditionally convergent series can change its sum.
When it fails
Take \(p=1\), the alternating harmonic series. It converges, because the signs make the teal dots approach a value near 0.69.
Make every term positive. The red bars turn green, and the teal dots rise along the orange dots without limit. A conditionally convergent series converges only because of its signs, so convergence does not imply absolute convergence.
Example 1: the magnitudes add up exactly
Classify \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2+n}\) as absolutely convergent, conditionally convergent or divergent.
Test the series of magnitudes first. Factor the denominator.
\[\cy{\sum\frac{1}{n^2+n}}=\sum\frac{1}{n(n+1)}\]Write each magnitude as a difference of two fractions. A common denominator confirms this.
\[\frac{1}{n(n+1)}=\frac1n-\frac{1}{n+1}\]Add the first N magnitudes. Each subtracted fraction cancels the fraction added after it, so only the first and the last remain.
\[\begin{aligned}&\left(1-\tfrac12\right)+\left(\tfrac12-\tfrac13\right)\\&+\cdots+\left(\tfrac1N-\tfrac1{N+1}\right)\\&=1-\frac{1}{N+1}\end{aligned}\]As N grows, this total approaches 1, as the orange dots show. The magnitudes converge, so the series is absolutely convergent.
\[\lim_{N\to\infty}\left(1-\frac{1}{N+1}\right)=1\]
Example 2: a converging series is not always absolute
Classify \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{2n+9}\).
Test the magnitudes. Compare \(\frac{1}{2n+9}\) with the harmonic series \(\sum\frac1n\). The ratio of the terms approaches \(\frac12\), a positive number.
\[\lim_{n\to\infty}\frac{1/(2n+9)}{1/n}=\frac12\]The harmonic series diverges, so the magnitudes diverge by the limit comparison test. The series is not absolutely convergent.
Test the original series. The signs alternate, and the sizes \(\frac{1}{2n+9}\) decrease to 0 because the denominator grows. By the alternating series test, the series converges, and the teal dots stay in the shaded strip.
\[\lim_{n\to\infty}\frac{1}{2n+9}=0\]A tempting wrong move: The series converges, so it is absolutely convergent.
\[\begin{aligned}\sum a_n\ &\text{converges}\\ \Rightarrow\ &\text{absolutely convergent}\end{aligned}\]Absolute convergence needs the orange dots to level off, but the red arrow shows that they continue to rise.
The series converges and its magnitudes diverge, so it is conditionally convergent.
Practice
Classify \(\displaystyle\sum_{n=1}^{\infty}(-1)^n\frac{3n}{n+2}\) as absolutely convergent, conditionally convergent or divergent.
Before any other test, find the size of the terms for large \(n\). Do the teal dots approach a value?
Solution
The sizes \(\frac{3n}{n+2}\) approach 3, as the dashed lines show.
\[\lim_{n\to\infty}\cy{\frac{3n}{n+2}}=3\]The terms alternate near 3 and −3. They do not approach 0, so the series diverges by the divergence test.
The series is divergent. It is neither absolutely nor conditionally convergent, because both of those names need the series to converge.
Questions
The graph shows the series \(1-\frac{1}{2^p}+\frac{1}{3^p}-\cdots\) from the opening, with the same bars and dots. It first shows \(p=3\), and question 2 changes it to \(p=0.8\).
In the first view, the orange dots level off at a limit. How is the series classified?
- Absolutely convergent
- Conditionally convergent
- Divergent
Answer
Absolutely convergent
In the second view, the orange dots rise without limit and the teal dots approach a value. How is the series classified?
- Conditionally convergent
- Absolutely convergent
- Divergent
Answer
Conditionally convergent
How is \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{3n^2+1}\) classified?
- Absolutely convergent
- Conditionally convergent
- Divergent
- It cannot be classified
Answer
Absolutely convergent
Summary
You test the magnitudes, then the original series if needed, and you name the result: absolutely convergent, conditionally convergent or divergent.
An absolutely convergent series converges whatever its signs, while a conditionally convergent series converges only because of its signs.
You use the names whenever a question asks how a series converges, and not only whether it converges.
This is one of four skills in topic 10.9 of Which & Why: AP Calculus.