Connect each conclusion to a valid test
Two series with negative terms
Series A is \(\sum(-1)^n\dfrac{3n}{5n+2}\), and series B is \(\sum\dfrac{\sin 4n}{n^{3/2}}\). Green bars are positive terms and red bars are negative terms. Drag the slider to add terms, and the teal dots show the running total.
In A, the signs alternate, but the sizes rise toward 0.6 instead of shrinking. The teal dots therefore alternate between values near \(-0.48\) and \(0.10\) and approach no single value.
In B, the signs follow no pattern, but the bars shrink quickly, so the teal dots approach a value near \(-0.5\).
The sign pattern alone decides nothing, because A has alternating signs and diverges while B has no pattern and converges.
Each test has conditions
The list below the graphs names three tests. The divergence test asks whether the terms fail to approach 0. The absolute-value test asks whether the series of magnitudes \(\sum|a_n|\), the sizes of the terms, converges. The alternating series test asks whether the signs alternate and the sizes decrease to 0.
Press each button to compare series A, series B and the alternating harmonic series. A test turns green when the series on the screen meets its conditions.
A conclusion is valid only when the conditions of its test hold, and the sign pattern alone meets the conditions of no test.
Notation
The divergence test says that if the terms do not approach 0, the series diverges.
\[\lim_{n\to\infty}a_n\ne0\ \Rightarrow\ \sum a_n\ \text{diverges}\]This test cannot show convergence. If the terms approach 0, it gives no answer.
The absolute-value test says that if the series of \(\cy{\text{magnitudes}}\) converges, the series converges. The comparison, p-series and ratio tests need positive terms, so you apply them to the magnitudes.
\[\begin{aligned}\cy{\sum|a_n|}\ &\text{converges}\\ \Rightarrow\ \sum a_n\ &\text{converges}\end{aligned}\]The alternating series test says that if the signs alternate and the sizes \(\cy{b_n}\) decrease to 0, the series converges.
\[\begin{aligned}&\sum(-1)^n\,\cy{b_n}:\\&b_{n+1}\le b_n,\quad b_n\to0\end{aligned}\]Write each conclusion with the name of the test and the conditions you checked.
Alternating signs alone prove nothing.
When it fails
Series A has alternating signs, so it is tempting to use the alternating series test. The test needs the sizes to decrease to 0, but the sizes increase toward \(\frac35\).
The teal dots move up and down and approach no value. The terms do not approach 0, so the divergence test shows that A diverges.
Example 1: the ratio test on a series with alternating signs
Decide whether \(\displaystyle\sum_{n=1}^{\infty}(-1)^n\frac{n^2}{2^n}\) converges.
The power \(2^n\) suits the ratio test, which uses the magnitudes, so the signs do not affect it.
Divide each magnitude by the one before it, and cancel the powers of 2. The orange dots show this ratio, and the faint dashed line marks 1.
\[\frac{(n+1)^2/2^{n+1}}{n^2/2^n}=\frac{(n+1)^2}{2n^2}\]As \(n\) grows, the ratio approaches \(\frac12\), the orange dashed line.
\[\lim_{n\to\infty}\frac{(n+1)^2}{2n^2}=\frac12\]The limit is less than 1, so the magnitudes converge by the ratio test. The series therefore converges absolutely.
Example 2: signs in pairs
The signs of a series come in pairs: \(1+\tfrac14-\tfrac19-\tfrac1{16}+\tfrac1{25}+\tfrac1{36}-\cdots\). The pattern continues, and the term with index \(n\) has size \(\frac{1}{n^2}\). Decide whether the series converges.
The sizes \(\frac{1}{n^2}\) shrink, and the signs come in pairs: two green, positive bars, then two red, negative bars.
A tempting wrong move: The sizes decrease and approach 0, so the alternating series test shows that the series converges.
\[b_n=\frac{1}{n^2}\downarrow0\ \Rightarrow\ \text{converges}\]The test needs the sign to change at every term. The ringed bars at \(n=1\) and \(n=2\) are both green, and those at \(n=3\) and \(n=4\) are both red.
Use the absolute-value test instead. The magnitudes \(\frac{1}{n^2}\) form a p-series with \(p=2>1\), so they converge.
\[\cy{\sum\frac{1}{n^2}}\ \text{converges}\]The magnitudes converge, so the series converges absolutely, whatever its sign pattern.
Practice
Decide whether \(\displaystyle\sum_{n=1}^{\infty}(-1)^n\frac{5}{2n+3}\) converges absolutely, converges conditionally or diverges. Name each test you use.
Test the terms first and the magnitudes second. The teal dots show whether the original series approaches one value.
Solution
The terms approach 0, so the divergence test gives no answer.
\[\lim_{n\to\infty}\frac{5}{2n+3}=0\]Compare the magnitudes with the harmonic series \(\sum\frac1n\), which diverges. The ratio of the terms approaches \(\frac52>0\), so the magnitudes diverge by the limit comparison test.
\[\lim_{n\to\infty}\frac{5/(2n+3)}{1/n}=\frac52\]The signs alternate and the sizes decrease to 0, so the series converges by the alternating series test. Every teal dot after the eighth stays inside the shaded strip.
The series converges and its magnitudes diverge, so it is conditionally convergent.
Questions
Green bars are positive terms, red bars are negative terms, and the teal dots show the running total. The graph first shows \(a_n=(-1)^n\frac{n}{n+1}\), and question 2 changes it to sizes \(\frac{1}{n^2}\) with signs that repeat +, −, −.
In the first view, a student claims: "The signs alternate, so the series converges." What is wrong?
- The bars do not shrink to 0, so the alternating series test does not apply.
- The signs do not alternate.
- Series with negative terms never converge.
Answer
The bars do not shrink to 0, so the alternating series test does not apply.
In the second view, which test can show that the series converges?
- The absolute-value test, by comparing the magnitudes with a p-series.
- The alternating series test.
- The divergence test.
Answer
The absolute-value test, by comparing the magnitudes with a p-series.
A student writes: "The series \(\sum(-1)^n/(n^2+5)\) converges because it is smaller than \(\sum1/n^2\)." Which statement describes this reasoning?
- The conclusion is correct, but the comparison must be made on the magnitudes.
- The reasoning is valid, because every term is smaller than the matching term of the p-series.
- The conclusion is wrong, because the terms are smaller than those of a convergent p-series.
- The reasoning is valid, because the signs alternate.
Answer
The conclusion is correct, but the comparison must be made on the magnitudes.
Summary
You support each conclusion about a series with a named test whose conditions you checked.
This works because a test proves its conclusion only when its conditions hold, and alternating signs alone meet the conditions of no test.
For any series with negative terms, you test the terms first, then the magnitudes, then the alternating pattern.
This is one of four skills in topic 10.9 of Which & Why: AP Calculus.