Direct Comparison Test
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 5.4: “Comparison Tests” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/5-4-comparison-tests |
| Textbook used in class | Stewart, Calculus, Section 11.4: “The Comparison Tests” (Examples 1, 2, 3) |
Opening Scenario
You want to know whether a stack of blocks will fall over, but the exact height is hard to measure. You notice the stack is shorter than a nearby structure you know is stable. If something smaller is fine, yours must be fine too. If something shorter collapses, yours certainly will. That reasoning is the Direct Comparison Test.
Quick Reference
Direct Comparison Test. Suppose $a_n \geq 0$ and $b_n \geq 0$ for all $n$ sufficiently large.
- If $a_n \leq b_n$ and $\sum b_n$ converges, then $\sum a_n$ converges.
- If $a_n \geq b_n$ and $\sum b_n$ diverges, then $\sum a_n$ diverges.
Memory rule: bounded above by a convergent series means converge; bounded below by a divergent series means diverge. The other two combinations tell you nothing.
Standard comparison targets:
- $p$-series $\sum 1/n^p$: converges if $p > 1$, diverges if $p \leq 1$.
- Geometric series $\sum r^n$: converges if $|r| < 1$, diverges if $|r| \geq 1$.
Key Concepts
1. The Logic of Comparison
If every term of $\sum a_n$ is at most the corresponding term of $\sum b_n$, then the partial sums of $\sum a_n$ are at most those of $\sum b_n$. If $\sum b_n$ is bounded (converges), so are the partial sums of $\sum a_n$; since $a_n \geq 0$, the partial sums are increasing and bounded, so $\sum a_n$ converges.
The divergence direction: if $a_n \geq b_n$ and $\sum b_n$ diverges (its partial sums grow without bound), the partial sums of $\sum a_n$ are even larger, so they also grow without bound.
2. Choosing the Comparison Series
For a rational expression in $n$, keep only the highest-power terms in the numerator and denominator to guess the comparison series. Then verify the inequality.
Example 1. Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{1}{2^n + 3}$ converges. (Stewart 11.4, Example 1.)
Idea. For large $n$, $2^n + 3 > 2^n$, so $\dfrac{1}{2^n + 3} < \dfrac{1}{2^n}$. The geometric series $\sum (1/2)^n$ converges, so try comparing to it.
Verification. Since $2^n + 3 > 2^n$ for all $n \geq 1$, we have $0 < \dfrac{1}{2^n + 3} < \dfrac{1}{2^n}$. The series $\sum 1/2^n$ is geometric with ratio $1/2 < 1$, so it converges. By the Direct Comparison Test, $\sum \dfrac{1}{2^n + 3}$ converges.
Boxed answer: Converges, by comparison to the convergent geometric series $\sum (1/2)^n$.
Example 2. Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{\ln n}{n}$ converges. (Stewart 11.4, Example 2.)
Idea. For $n \geq 3$, $\ln n > 1$, so $\dfrac{\ln n}{n} > \dfrac{1}{n}$. The harmonic series $\sum 1/n$ diverges.
Verification. For $n \geq 3$, $\ln n \geq 1$, so $\dfrac{\ln n}{n} \geq \dfrac{1}{n}$. Since $\sum 1/n$ diverges, the Direct Comparison Test gives: $\sum \dfrac{\ln n}{n}$ diverges.
Boxed answer: Diverges, by comparison to the divergent harmonic series $\sum 1/n$.
3. What the Test Cannot Do
The comparison runs in only two useful directions. If you find $a_n \leq b_n$ but $\sum b_n$ diverges, that is no help: being smaller than a divergent series does not imply convergence. Similarly, if $a_n \geq b_n$ but $\sum b_n$ converges, being larger than a convergent series does not imply divergence.
trying the “wrong direction” comparison. A student who wants to show $\sum \frac{1}{n^2 + n}$ converges might note that $\frac{1}{n^2 + n} < \frac{1}{n^2}$, which is the correct direction (bounded above by the convergent $p$-series $\sum 1/n^2$, so it converges). But if that student instead noted $\frac{1}{n^2 + n} > \frac{1}{n^3}$ and compared to $\sum 1/n^3$ (which also converges), that comparison goes in the wrong direction for the test. Being larger than a convergent series says nothing. The inequality must put the unknown series on top of a convergent one, or on the bottom of a divergent one.
Common Errors Summary
| Error | Correction |
|---|---|
| Using the test when $a_n \leq b_n$ but $\sum b_n$ diverges | That direction is inconclusive; try a smaller comparison or use another test |
| Failing to verify the inequality (just guessing) | Explicitly confirm $a_n \leq b_n$ for all $n$ beyond some index |
| Comparing to $\sum 1/n^p$ with the wrong $p$ | The dominant term determines the appropriate $p$; check the inequality direction |
Common Misconceptions
if $a_n \le b_n$ and $\sum b_n$ diverges, then $\sum a_n$ diverges.
This is the concept-image-conflicts-definition error about which direction of comparison is valid. Being bounded above by a divergent series gives no information: the smaller series could still converge. The valid conclusions are only that a series bounded above by a convergent series converges, or a series bounded below by a divergent series diverges. The series $\sum 1/n^2$ satisfies $1/n^2 < 1/n$ and $\sum 1/n$ diverges, yet $\sum 1/n^2$ converges.
the comparison inequality is valid in either direction for proving convergence.
This is the concept-image-conflicts-definition error about the direction of the bound. To prove $\sum a_n$ converges, the series must be bounded above by a convergent series: $a_n \le b_n$ with $\sum b_n$ convergent. Finding that $a_n \ge b_n$ where $\sum b_n$ converges says nothing about $\sum a_n$; being larger than a convergent series does not imply convergence or divergence.
Leveled Practice
Level 1
Problem 1. Does $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2 + n + 1}$ converge?
Show answer
For $n \geq 1$, $n^2 + n + 1 > n^2$, so $\dfrac{1}{n^2 + n + 1} < \dfrac{1}{n^2}$. Since $\sum 1/n^2$ converges ($p$-series, $p = 2 > 1$), the Direct Comparison Test gives convergence.
Problem 2. Does $\displaystyle\sum_{n=1}^{\infty} \frac{1}{\sqrt{n} - 1}$ converge (for $n \geq 2$)?
Show answer
For $n \geq 2$, $\sqrt{n} - 1 < \sqrt{n}$, so $\dfrac{1}{\sqrt{n} - 1} > \dfrac{1}{\sqrt{n}} = \dfrac{1}{n^{1/2}}$. Since $\sum 1/n^{1/2}$ is a $p$-series with $p = 1/2 \leq 1$, it diverges. The series $\sum \dfrac{1}{\sqrt{n}-1}$ diverges.
Level 2
Problem 3. Does $\displaystyle\sum_{n=1}^{\infty} \frac{n+1}{n^3 + 2}$ converge?
Show answer
The dominant behavior is $n/n^3 = 1/n^2$. Check: $\dfrac{n+1}{n^3 + 2} \leq \dfrac{n + n}{n^3} = \dfrac{2n}{n^3} = \dfrac{2}{n^2}$ for $n \geq 2$. Since $\sum 2/n^2 = 2\sum 1/n^2$ converges, the series converges.