Test the absolute-value series

Two totals for the same terms

The series is \(1-\tfrac18+\tfrac1{27}-\tfrac1{64}+\cdots\), so its terms have size \(\frac{1}{n^3}\) and their signs alternate. The lower graph shows the terms: green bars are positive and red bars are negative.

Move the slider to add terms. In the upper graph, the teal dots add the terms with their signs, and the orange dots add their sizes.

After four terms, the teal total is 0.896, less than the orange total of 1.178, because the negative terms subtract.

As you add more terms, the orange total rises by less each time and approaches a limit near 1.202, shown by the dashed line. The teal total stays below it.

Any series with both signs

The same scene works for any series with terms \(a_n\), whose signs can follow any pattern, as in the lower graph.

The size \(|a_n|\) of a term is its magnitude. The series of magnitudes adds these sizes, and the orange dots show its running total.

Press the button to split the terms into two groups. The green dots add the positive terms, and the red dots add the sizes of the negative terms.

Each group is part of the orange total, so if the orange total has a limit, neither group can pass it. A group total only grows, so it then approaches a value.

The teal total is the green total minus the red total, so it approaches a value too.

Notation

1. The series of magnitudes adds the sizes \(|a_n|\). Its terms are never negative, so the tests for positive terms, such as the comparison test, apply to it.

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

2. If the series of magnitudes converges, the original series converges, because the two group totals then approach values and so does their difference. The dashed line marks the limit of the orange total.

   \[\begin{aligned}\cy{\sum|a_n|}\ &\text{converges}\\ \Rightarrow\ \cb{\sum a_n}\ &\text{converges}\end{aligned}\]

3. A series whose magnitudes converge is called absolutely convergent.

4. The rule works in one direction only. When the series of magnitudes diverges, the rule gives no answer.

   The original series might still converge.

When it fails

With the exponent 1, the terms are \(1-\tfrac12+\tfrac13-\cdots\). The orange dots then add the harmonic series \(1+\tfrac12+\tfrac13+\cdots\), which diverges, so they rise without limit.

The teal dots still approach a value near 0.69, so a series can converge while its magnitudes diverge.

Example 1: signs that alternate

Decide whether \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2+1}\) converges.

1. Write the series of magnitudes. The sign factor \((-1)^n\) has size 1, so it disappears.

   \[\cy{\sum\frac{1}{n^2+1}}\]

2. The denominator \(n^2+1\) is larger than \(n^2\), so each magnitude is smaller than the term of the p-series \(\sum\frac{1}{n^2}\). The faint dots show the running total of this p-series, and the orange dots stay below them.

   \[\frac{1}{n^2+1}<\frac{1}{n^2}\]

3. This p-series has \(p=2>1\), so it converges. By the comparison test, the series of magnitudes converges.

4. Therefore the original series converges. It is absolutely convergent.

Example 2: signs with no pattern

Decide whether \(\displaystyle\sum_{n=1}^{\infty}\frac{\cos n}{n^2}\) converges.

1. The signs of \(\cos n\) follow no pattern: \(\cos 1\) is positive, while \(\cos 2\), \(\cos 3\) and \(\cos 4\) are negative. The alternating series test does not apply, so test the series of magnitudes.

2. A tempting wrong move: Because \(\cos n\le 1\), each term is at most \(\frac{1}{n^2}\). Compare with the convergent p-series \(\sum\frac{1}{n^2}\).

   \[\frac{\cos n}{n^2}\le\frac{1}{n^2}\]

   The dots at \(\frac{1}{n^2}\), now red, lie above the axis, but the red bars of the negative terms lie below it. This bound does not limit how far down those bars reach, and the comparison test needs positive terms.

3. Compare the magnitudes instead. Because \(|\cos n|\le1\), each magnitude is at most \(\frac{1}{n^2}\), so every bar lies between the faint dots at \(\frac{1}{n^2}\) and \(-\frac{1}{n^2}\).

   \[\cy{\left|\frac{\cos n}{n^2}\right|}\le\frac{1}{n^2}\]

4. Because \(2>1\), the p-series \(\sum\frac{1}{n^2}\) converges. By comparison, the magnitudes converge, so the original series converges.

Practice

Decide whether \(\displaystyle\sum_{n=1}^{\infty}\frac{(-2)^n}{3^n+n}\) converges.

Write the series of magnitudes. Find a series with known behavior whose terms are larger than each magnitude.

Questions

The bars show the terms of an alternating series: green bars are positive and red bars are negative. The orange dots show the running total of the sizes, and the teal dots show the running total with signs. The graph first shows sizes \(\frac{1}{n^2}\), and question 2 changes them to \(\frac{1}{\sqrt n}\).

1. In the first view, the orange dots level off at a limit. What follows?

   • The magnitudes converge, so the original series converges.
   • The original series converges to the height of the orange dots.
   • The original series might diverge, because it has negative terms.
   Answer

   The magnitudes converge, so the original series converges.

2. In the second view, the orange dots rise without limit. What follows?

   • The rule gives no answer, so the original series needs another test.
   • The original series diverges.
   • The original series converges, because the teal dots stay close together.
   • The original series converges absolutely.
   Answer

   The rule gives no answer, so the original series needs another test.

3. For which series do the magnitudes show that the series converges?

   • \(\displaystyle\sum\frac{\sin 2n}{n^2}\)
   • \(\displaystyle\sum\frac{(-1)^n}{\sqrt n}\)
   • \(\displaystyle\sum(-1)^n\frac{n}{n+1}\)
   Answer

   \(\displaystyle\sum\frac{\sin 2n}{n^2}\)

Summary

You replace each term by its size and test that series of positive terms with a comparison test or a p-series test.

If the sizes have a finite total, the positive terms and the negative terms each have a finite total, so the series converges.

This test comes first for a series with negative terms, because a finite total of sizes proves convergence while a divergent total decides nothing.

This is one of four skills in topic 10.9 of Which & Why: AP Calculus.