Recognize an inconclusive comparison
Two series under the same divergent series
The chalk outlines are the terms \(b_n=\frac{1}{\sqrt{n}}\), a p-series with \(p=\frac12\), so it diverges. The orange bars are the terms \(a_n=\frac{1}{n^{2}}\), and none is taller than the chalk bar behind it.
The middle graph shows the partial sums. The chalk curve rises without bound, and the teal curve, the orange total, approaches a limit, so this orange series converges.
Turn on the toggle. The orange bars are now \(\frac1n\), still no taller than the chalk bars, but their total rises without bound, so this orange series diverges.
Both orange series lie below the same divergent series, and for both the ratio in the bottom graph goes to 0. The inequality and the limit are the same, but one series converges and the other diverges, so these facts do not decide the question.
Four arrangements, two that decide
These graphs have no numbers. The orange series is the one to test, the chalk series is one that you know, and all terms are positive.
The orange bars can lie below or above the chalk bars, and the chalk series can converge or diverge, which gives four arrangements.
Press the first two buttons. Below a convergent series, the orange total has a bound, so the orange series converges. Above a divergent series, it has no bound, so the orange series diverges.
Press the other four buttons. Below a divergent series or above a convergent one, the orange total can stay bounded or rise without bound, so these two arrangements decide nothing.
Notation
Direct comparison with \(0\le a_n\le b_n\) decides only when \(\sum b_n\) converges.
\[\begin{aligned}&\sum b_n\ \text{converges}\\ &\quad\Rightarrow\ \sum a_n\ \text{converges}\\ &\sum b_n\ \text{diverges}\\ &\quad\Rightarrow\ \text{no conclusion}\end{aligned}\]Direct comparison with \(a_n\ge b_n\ge 0\) decides only when \(\sum b_n\) diverges.
\[\begin{aligned}&\sum b_n\ \text{diverges}\\ &\quad\Rightarrow\ \sum a_n\ \text{diverges}\\ &\sum b_n\ \text{converges}\\ &\quad\Rightarrow\ \text{no conclusion}\end{aligned}\]Limit comparison decides when the limit \(c\) is finite and greater than 0. If \(c=0\) or \(c\) is infinite, the two series need not behave in the same way.
\[\begin{aligned}\lim\frac{a_n}{b_n}&=0\ \text{or}\ \infty\\ \Rightarrow\ &\text{the test does not decide}\end{aligned}\]A limit of 0 can still help when \(\sum b_n\) converges, and an infinite limit can help when \(\sum b_n\) diverges. The AP test names only a finite \(c\) greater than 0.
When a comparison is inconclusive, choose a different series \(b_n\), and compare again.
When it fails
Turn on the toggle to test \(\frac{\cos n}{n^{2}}\) against \(\frac{1}{n^{2}}\). Some of its terms are negative, so their bars are red and lie below the axis.
Both comparison tests need positive terms, so neither test applies here, whatever the other series does.
Example 1: The inequality points the wrong way
Decide whether \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{3^{n}-2}\) converges.
Compare with the convergent geometric series \(\frac{1}{3^{n}}\). The bottom \(3^{n}-2\) is smaller, so each term is larger.
\[\frac{1}{3^{n}-2}>\frac{1}{3^{n}}\]The orange bars lie above a convergent series, so this comparison is inconclusive.
Try the limit comparison with the same series. Divide the top and the bottom by \(3^{n}\).
\[\frac{a_n}{b_n}=\frac{3^{n}}{3^{n}-2}=\frac{1}{1-2/3^{n}}\to\cy{1}\]The limit is finite and greater than 0, and the geometric series converges, so the given series converges.
Example 2: A limit of 0
A student compares \(\displaystyle\sum_{n=1}^{\infty}\frac{n+1}{n^{4}}\) with \(\displaystyle\sum_{n=1}^{\infty}\frac1n\). Decide whether the first series converges.
Form the ratio. Its bottom \(n^{3}\) grows faster than its top \(n+1\), so the ratio goes to 0.
\[\frac{a_n}{b_n}=\frac{n+1}{n^{3}}\to\cy{0}\]A tempting wrong move: The limit exists, and the series of \(\frac1n\) diverges, so the first series diverges.
\[\begin{aligned}\sum\frac1n\ \text{diverges}&\ \Rightarrow\\ \sum\frac{n+1}{n^{4}}&\ \text{diverges}\end{aligned}\]A limit of 0 means that the orange terms are much smaller than the chalk terms, and terms below a divergent series decide nothing. The teal total stays below the dashed line at 3.
Compare with a convergent series that lies above. For \(n\ge 1\), \(n+1\le 2n\), so each term is at most \(\frac{2}{n^{3}}\), shown by the faint bars.
\[\frac{n+1}{n^{4}}\le\frac{2n}{n^{4}}=\cq{\frac{2}{n^{3}}}\]The series of \(\frac{2}{n^{3}}\) is a constant times a p-series with \(p=3\), so it converges. The first series lies below it, so it converges.
Practice
For each case, say whether the comparison decides the first series. (a) \(\frac{1}{n-1}\) for \(n\ge2\), with \(\frac1n\). (b) \(\frac{1}{n^{2}+n}\), with \(\frac1n\). (c) \(\frac{\cos n}{n^{2}}\), with \(\frac{1}{n^{2}}\).
Ask three questions. Are the terms positive? Do they lie above or below the second series? Does the second series converge or diverge?
Solution
Case (a). The bottom \(n-1\) is smaller than \(n\), so the terms lie above a divergent series. The first series diverges.
\[\frac{1}{n-1}>\frac1n,\quad \sum\frac1n\ \text{diverges}\]Case (b). The bottom \(n^{2}+n\) is larger than \(n\), so the terms lie below a divergent series. This is inconclusive.
\[\frac{1}{n^{2}+n}<\frac1n,\quad \sum\frac1n\ \text{diverges}\]Case (c). Some terms are negative, so neither test applies.
Questions
Questions 1 and 2 use the figure. The orange bars belong to a series to test, the chalk outlines to a series that you know, and all terms are positive. The bottom graph shows the ratio of the middle graph's orange term to its chalk term.
In the top graph, the orange bars lie above the chalk bars of \(\frac{1}{n^{2}}\), and that series converges. What can you conclude from this comparison?
- Nothing
- The orange series converges
- The orange series diverges, because it lies above
Answer
Nothing
The dots in the bottom graph go to 0, and the chalk series in the middle graph diverges. What does the limit comparison show about the orange series?
- Nothing
- It converges, because the ratio goes to 0
- It diverges, because the chalk series diverges
Answer
Nothing
Which comparison shows that \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{3}+n}\) converges?
- \(\frac{1}{n^{3}+n}<\frac{1}{n^{3}}\), and the series of \(\frac{1}{n^{3}}\) converges
- \(\frac{1}{n^{3}+n}\ge\frac{1}{2n^{3}}\), and the series of \(\frac{1}{2n^{3}}\) converges
- \(\frac{1}{n^{3}+n}<\frac1n\), and the series of \(\frac1n\) diverges
- The ratio to \(\frac1n\) goes to 0, and the series of \(\frac1n\) diverges
Answer
\(\frac{1}{n^{3}+n}<\frac{1}{n^{3}}\), and the series of \(\frac{1}{n^{3}}\) converges
Summary
Before you conclude, you check which way the inequality points, whether the known series converges, and whether the limit is finite and greater than 0.
A direct comparison decides only when the tested series lies below a convergent series or above a divergent one.
When a comparison is inconclusive, you choose a different known series and compare again.
This is one of four skills in topic 10.6 of Which & Why: AP Calculus.