Use direct comparison in the correct direction
A smaller series under a larger one
The chalk bars are the terms \(b_n=\frac{1}{n^{3}}\), which form a p-series with \(p=3\) that converges. The orange bars are the terms \(a_n=\frac{1}{n^{3}+4}\), and each is shorter than the chalk bar around it, because its bottom is larger.
The lower graph shows the partial sums, the totals of the first N terms, with teal for the orange series. Drag N. After 10 terms the orange total is about 0.35 and the chalk total is about 1.20.
All the terms are positive, so each total only rises. The chalk total stays below the dashed line at the sum of the chalk series, because that series converges. The orange total is lower, so it stays below the line too.
A total that rises and stays below a fixed number approaches a limit, so the orange series converges.
Two arrangements that decide a series
The orange series is the one to test, and the chalk series is one that you know. All the terms are positive.
With the toggle off, the orange bars lie below the chalk bars, and the chalk total stays below the dashed line. The orange total is smaller, so it has the same bound, and it approaches a limit.
Turn the toggle on. The orange bars lie above the chalk bars, and the chalk total rises past every height. The orange total is larger, so it rises past every height too.
A bound from above can show convergence, and a bound from below can show divergence.
Notation
Let \(a_n\) be the terms to test, and \(b_n\) the terms of a series that you know. Both must be at least 0.
\[\cy{a_n}\ge 0,\qquad b_n\ge 0\]The terms \(b_n\) are an upper bound if \(a_n\) is at most \(b_n\). If the series of an upper bound converges, the series of \(a_n\) converges.
\[\begin{aligned}\cy{a_n}\le b_n,\ \ \sum b_n&\ \text{converges}\\ \Rightarrow\ \sum a_n&\ \text{converges}\end{aligned}\]The terms \(b_n\) are a lower bound if \(a_n\) is at least \(b_n\). If the series of a lower bound diverges, the series of \(a_n\) diverges.
\[\begin{aligned}\cy{a_n}\ge b_n,\ \ \sum b_n&\ \text{diverges}\\ \Rightarrow\ \sum a_n&\ \text{diverges}\end{aligned}\]The inequality needs to hold only from some n on, because finitely many terms do not change whether a series converges.
A fraction becomes smaller when its bottom becomes larger, and larger when its top becomes larger. Use this to find which way an inequality points.
\[\cy{\frac{1}{n^{3}+4}}<\frac{1}{n^{3}}\]
When it fails
Take \(a_n=-1\) for every \(n\), and \(b_n=\frac{1}{n^{2}}\). Then \(a_n\) is at most \(b_n\), and the series of \(b_n\) converges.
The red bars show that every term is negative. The teal total stays below the chalk total but decreases without bound, so the series diverges. The rule needs \(a_n\ge 0\), because only a rising total below a bound must approach a limit.
Example 1: Below a geometric series
Decide whether \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{4^{n}+n}\) converges.
The bottom \(4^{n}+n\) is larger than \(4^{n}\), so each term is smaller than \(\frac{1}{4^{n}}\), the term of a geometric series. The chalk bars show that series.
\[0<\cy{\frac{1}{4^{n}+n}}<\frac{1}{4^{n}}\]The geometric series has the ratio \(\frac14\), which is less than 1, so it converges. Its sum is the first term divided by 1 minus the ratio.
\[\sum_{n=1}^{\infty}\left(\frac14\right)^{n}=\frac{1/4}{1-1/4}=\frac13\]The terms are positive, so the teal total rises. It stays below the chalk total, which approaches \(\frac13\), so the series converges.
Example 2: Above a divergent series
Decide whether \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{n+\sqrt{n}}\) diverges.
The bottom \(n+\sqrt{n}\) is larger than \(n\), so each term is smaller than \(\frac1n\). The chalk outlines show the terms \(\frac1n\).
A tempting wrong move: The terms are smaller than \(\frac1n\), and the series of \(\frac1n\) diverges, so this series diverges.
\[\begin{aligned}a_n<\frac1n,\ \ \sum\frac1n&\ \text{diverges}\\ \Rightarrow\ \sum a_n&\ \text{diverges}\end{aligned}\]Lying below a divergent series proves nothing. The red dots show the terms \(\frac{1}{n^{2}}\), which are also smaller than \(\frac1n\), but their series converges.
Look for a divergent series below the terms instead. For \(n\ge 1\), \(\sqrt{n}\le n\), so the bottom is at most \(2n\). The faint outlines show the terms \(\frac{1}{2n}\).
\[n+\sqrt{n}\le 2n\ \Rightarrow\ \cy{\frac{1}{n+\sqrt{n}}}\ge\cq{\frac{1}{2n}}\]The series of \(\frac{1}{2n}\) is \(\frac12\) times the harmonic series, so it diverges. The orange bars lie above the faint outlines, so the orange series diverges.
Practice
Decide whether \(\displaystyle\sum_{n=1}^{\infty}\frac{\sqrt[3]{n}}{n^{3}+1}\) converges.
Remove the 1 from the bottom, and write the result as a power of \(n\). Move the slider until the chalk bars lie just above the orange bars.
Solution
Removing the 1 makes the bottom smaller, so the fraction becomes larger, and the result is one power of \(n\).
\[\cy{\frac{\sqrt[3]{n}}{n^{3}+1}}<\frac{n^{1/3}}{n^{3}}=\frac{1}{n^{8/3}}\]This is a p-series with \(p=\frac83\), which is greater than 1, so it converges. The terms are positive and lie below it, so the series converges.
Questions
Questions 1 and 2 use the figure. The orange bars belong to a series to test, and the chalk outlines belong to a series that you know converges or diverges.
In the upper graph, the orange bars lie below the chalk bars of \(\frac{1}{n^{2}}\). What follows for the orange series?
- It converges
- It diverges
- Nothing follows
Answer
It converges
In the lower graph, the orange bars lie above the chalk bars of \(\frac1n\). What follows for the orange series?
- It diverges
- It converges, because the orange bars shrink
- Nothing follows
Answer
It diverges
What does a comparison show about \(\displaystyle\sum_{n=1}^{\infty}\frac{\cos^{2}n}{2^{n}}\)?
- It converges, because \(0\le\frac{\cos^{2}n}{2^{n}}\le\frac{1}{2^{n}}\) and the series of \(\frac{1}{2^{n}}\) converges
- Nothing, because \(\cos n\) changes sign
- Nothing, because the inequality holds only for some n
Answer
It converges, because \(0\le\frac{\cos^{2}n}{2^{n}}\le\frac{1}{2^{n}}\) and the series of \(\frac{1}{2^{n}}\) converges
Summary
You compare a series whose terms are at least 0 with a series that you know converges or diverges.
A rising total below a total with a bound approaches a limit, and a total above a total with no bound has no bound.
You use an upper bound to show convergence and a lower bound to show divergence.
This is one of four skills in topic 10.6 of Which & Why: AP Calculus.