In the first view, the bars shrink toward 0 and the signs alternate. What does the alternating series test give?

The series diverges, because the sizes add up without limit.

The series converges absolutely.

In the second view, the signs alternate. Can the alternating series test show convergence?

No, because the sizes do not approach 0.

Yes, because the signs alternate.

Yes, because the teal dots stay between two heights.

Topic 10.9 · BC

Test the original series when necessary

Builds on testing the absolute-value series. 7 minutes.

Sizes that add up without limit

The series is \(1-\tfrac13+\tfrac15-\tfrac17+\cdots\), with general term \(a_n=\dfrac{(-1)^{n+1}}{2n-1}\). The lower graph shows the terms, green when positive and red when negative.

The slider adds terms. The orange dots show the running total of the sizes. It rises without limit, because each size is larger than \(\frac{1}{2n}\) and these add up to half the harmonic series.

The series therefore does not converge absolutely, but the teal dots, which show the running total with the signs, may still approach a value.

The teal total moves up and down, from 1 to 0.667 to 0.867 to 0.724, and each move is shorter than the one before.

Press the button. The teal dots approach the dashed line near 0.785, although the orange total has no limit.

Any alternating series

For any alternating series, write the size of term \(n\) as \(b_n\), so the series is \(b_1-b_2+b_3-\cdots\).

Each term moves the teal dot by its size, in the opposite direction to the move before. The orange bracket shows the move that the slider chooses.

If each size is smaller than the one before, each move undoes only part of the move before it. Shade the strip between two dots: every later dot lies inside it.

If the sizes also approach 0, the strip narrows and the dots approach one value, even if the sizes add up without limit.

Notation

First test the series of magnitudes, the sizes \(|a_n|\). If it diverges, this test gives no answer, so return to the original series.
\[\begin{aligned}\cy{\sum|a_n|}\ &\text{diverges}\\ \Rightarrow\ \text{test}\ \cb{\sum a_n}\ &\text{itself}\end{aligned}\]
If the signs alternate, write the series with positive sizes \(b_n\). The orange bracket beside a bar shows its size, whether the bar is green or red.
\[\sum(-1)^{n+1}\,\cy{b_n},\qquad b_n>0\]
The alternating series test has two conditions: each size is no larger than the one before, and the sizes approach 0.
\[b_{n+1}\le b_n,\qquad \lim_{n\to\infty}b_n=0\]
The sizes need to decrease only from some term onward.
If both conditions hold, the series converges, because every later dot stays inside the strip between two earlier dots. If the magnitudes also diverge, the convergence is conditional: the series converges, but not absolutely.

When it fails

Change the sizes to \(b_n=\dfrac{2n-1}{2n+1}\), which are \(\tfrac13,\tfrac35,\tfrac57,\dots\). These sizes increase toward 1, so they do not approach 0.

Each move is longer than the one before, so the teal dots do not approach a value. The terms do not approach 0, so the series diverges.

Example 1: the magnitudes diverge

Decide whether \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{3n+1}\) converges. If it does, say whether it converges absolutely.

Compare the magnitudes \(\frac{1}{3n+1}\) with the harmonic series \(\sum\frac1n\). The ratio of the terms approaches \(\frac13>0\), so the orange dots rise like the faint dots, a third of the harmonic total.
\[\lim_{n\to\infty}\frac{1/(3n+1)}{1/n}=\frac13\]
The harmonic series diverges, so the magnitudes diverge by the limit comparison test. This gives no answer about the original series.
Return to the original series. Its signs alternate, and the sizes \(\frac{1}{3n+1}\) decrease to 0 because \(3n+1\) grows without limit.
\[b_{n+1}
The alternating series test shows convergence. The magnitudes diverge, so the convergence is conditional.

Example 2: divergent magnitudes, convergent series

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

Test the magnitudes. For \(n\ge1\), \(n+3\le 4n\), so \(\sqrt{n+3}\le 2\sqrt n\), and reciprocals reverse the inequality.
\[\frac{1}{\sqrt{n+3}}\ge\frac{1}{2\sqrt n}\]
The series \(\sum\frac{1}{2\sqrt n}\) is a constant times a p-series with \(p=\frac12\), so it diverges. By comparison, the magnitudes diverge.
A tempting wrong move: The magnitudes diverge, so the original series diverges.
\[\begin{aligned}\sum\frac{1}{\sqrt{n+3}}\ &\text{diverges}\\ \Rightarrow\ \sum\frac{(-1)^n}{\sqrt{n+3}}\ &\text{diverges}\end{aligned}\]
The orange dots rise without limit, but the teal dots after the eighth stay inside the strip because each move is shorter than the last. Divergent magnitudes give no answer.
The original series has alternating signs, and its sizes \(\frac{1}{\sqrt{n+3}}\) decrease to 0, so the alternating series test shows convergence. The magnitudes diverge, so the convergence is conditional.
\[\lim_{n\to\infty}\cy{\frac{1}{\sqrt{n+3}}}=0\]

Practice

Decide whether \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{\sqrt[3]{n}}\) converges. If it does, say whether it converges absolutely.

Compare the orange dots with the teal dots. If the magnitudes diverge, test the original series.

Solution

The magnitudes \(\frac{1}{n^{1/3}}\) form a p-series with \(p=\frac13\le1\), so they diverge, and the orange dots in the upper graph rise without limit.
\[\cy{\sum\frac{1}{n^{1/3}}}\]
The signs alternate, and the sizes \(n^{-1/3}\) decrease to 0, so the alternating series test shows convergence. The teal dots in the lower graph narrow toward one value.
\[\lim_{n\to\infty}\cy{n^{-1/3}}=0\]
The magnitudes diverge, so the series converges conditionally.

Questions

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

In the first view, the bars shrink toward 0 and the signs alternate. What does the alternating series test give?
- The series converges.
- The series diverges, because the sizes add up without limit.
- The series converges absolutely.
Answer
The series converges.
In the second view, the signs alternate. Can the alternating series test show convergence?
- No, because the sizes do not approach 0.
- Yes, because the signs alternate.
- Yes, because the teal dots stay between two heights.
Answer
No, because the sizes do not approach 0.
Which series converges conditionally?
- \(\displaystyle\sum\frac{(-1)^n}{5n+2}\)
- \(\displaystyle\sum\frac{(-1)^n}{n^2}\)
- \(\displaystyle\sum(-1)^n\frac{5n}{5n+2}\)
Answer
\(\displaystyle\sum\frac{(-1)^n}{5n+2}\)

Summary

When the magnitudes diverge, you return to the original series and apply the alternating series test.

This works because moves that alternate in direction and decrease to 0 keep each new running total between the two before it.

If the test shows convergence, the series converges conditionally, because its series of magnitudes diverges.

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