Limit 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 4, 5) |
Opening Scenario
The Direct Comparison Test requires you to verify an explicit inequality between terms. That is sometimes awkward: the terms of a complicated series may not be obviously above or below a familiar series. The Limit Comparison Test sidesteps the inequality: if $a_n$ and $b_n$ behave the same way as $n \to \infty$ (their ratio tends to a finite positive limit), the two series either both converge or both diverge.
Quick Reference
Limit Comparison Test. Suppose $a_n > 0$ and $b_n > 0$ for all $n$ sufficiently large, and suppose \[ \lim_{n \to \infty} \frac{a_n}{b_n} = L. \]
- If $0 < L < \infty$: $\sum a_n$ and $\sum b_n$ converge or diverge together.
- If $L = 0$ and $\sum b_n$ converges: $\sum a_n$ converges.
- If $L = \infty$ and $\sum b_n$ diverges: $\sum a_n$ diverges.
The most useful case is $0 < L < \infty$. The cases $L = 0$ or $L = \infty$ are edge cases that reduce to one direction each.
Choosing $b_n$: For a rational expression, keep only the leading-power terms in numerator and denominator, then simplify. That simplified expression is the comparison series.
Key Concepts
1. Why the Limit Condition Is Enough
If $\lim a_n/b_n = L$ with $L > 0$, then for large $n$, $a_n \approx L \cdot b_n$. Multiplying all terms by the constant $L$ does not change convergence. So the two series have the same long-term behavior.
2. Applying the Test
Example 1. Does $\displaystyle\sum_{n=1}^{\infty} \frac{2n^2 + 3n}{\sqrt{5 + n^5}}$ converge? (Stewart 11.4, Example 4.)
Choose $b_n$. The numerator is dominated by $2n^2$ and the denominator by $\sqrt{n^5} = n^{5/2}$. So try $b_n = n^2 / n^{5/2} = n^{-1/2} = 1/\sqrt{n}$.
Compute the limit. \[ \frac{a_n}{b_n} = \frac{2n^2 + 3n}{\sqrt{5 + n^5}} \cdot \sqrt{n} = \frac{(2n^2 + 3n)\sqrt{n}}{\sqrt{5 + n^5}}. \] Divide numerator and denominator by $n^{5/2}$: \[ = \frac{2 + 3/n}{\sqrt{5/n^5 + 1}} \to \frac{2}{\sqrt{1}} = 2 \quad \text{as } n \to \infty. \]
Since $L = 2 \in (0, \infty)$, the series $\sum a_n$ behaves like $\sum 1/\sqrt{n}$, which is a $p$-series with $p = 1/2 \leq 1$: diverges.
Boxed answer: Diverges.
Example 2. Does $\displaystyle\sum_{n=1}^{\infty} \frac{1}{2^n - 1}$ converge? (Stewart 11.4, Example 5.)
Choose $b_n$. For large $n$, $2^n - 1 \approx 2^n$, so try $b_n = 1/2^n$.
Compute the limit. \[ \frac{a_n}{b_n} = \frac{1}{2^n - 1} \cdot 2^n = \frac{2^n}{2^n - 1} = \frac{1}{1 - 1/2^n} \to 1. \]
Since $L = 1 \in (0, \infty)$ and $\sum 1/2^n$ is a convergent geometric series, $\sum \dfrac{1}{2^n - 1}$ converges.
Boxed answer: Converges.
3. When to Use Limit Comparison vs. Direct Comparison
Use the Limit Comparison Test when the terms resemble a familiar series but an explicit inequality is hard to write down. Use the Direct Comparison Test when the inequality is obvious and easy to verify. For polynomial or rational expressions in $n$, Limit Comparison is almost always faster.
using $L = 0$ or $L = \infty$ to conclude the opposite. If $L = 0$ and $\sum b_n$ diverges, that tells you nothing about $\sum a_n$. Similarly for $L = \infty$ and $\sum b_n$ converges. Only the cases listed in the Quick Reference above are valid conclusions.
Common Errors Summary
| Error | Correction |
|---|---|
| Choosing $b_n$ with the wrong dominant terms | Keep only the highest-degree terms in numerator and denominator |
| Not simplifying the limit before evaluating | Divide numerator and denominator by the dominant power |
| Applying the test when $a_n$ is not eventually positive | The test requires $a_n > 0$; for alternating series use the Alternating Series Test |
Common Misconceptions
if $\lim a_n/b_n = 0$ and $\sum b_n$ diverges, then $\sum a_n$ diverges.
This is the concept-image-conflicts-definition error about the edge cases of the limit comparison test. The case $L = 0$ with $\sum b_n$ divergent gives no information about $\sum a_n$. Only the case $0 < L < \infty$ gives the shared fate conclusion. The case $L = 0$ with $\sum b_n$ convergent does imply $\sum a_n$ converges, but the divergent-$b_n$ case is always inconclusive when $L = 0$.
the comparison series $b_n$ can be any series, not just one whose behavior is known.
This is the action-view-of-function error about the test’s requirements. The limit comparison test produces a useful conclusion only when the behavior of $\sum b_n$ is already established. The standard choices are $p$-series $\sum 1/n^p$ (convergent for $p > 1$, divergent for $p \le 1$) and geometric series. Choosing a $b_n$ whose convergence is itself unknown leaves the comparison circular.
Leveled Practice
Level 1
Problem 1. Does $\displaystyle\sum_{n=1}^{\infty} \frac{n^2 + 1}{n^4 + 2}$ converge?
Show answer
Leading terms: $n^2/n^4 = 1/n^2$. Compute $\lim \dfrac{(n^2+1)/(n^4+2)}{1/n^2} = \lim \dfrac{n^4 + n^2}{n^4 + 2} = 1$. Since $\sum 1/n^2$ converges ($p = 2$), the series converges.
Problem 2. Does $\displaystyle\sum_{n=1}^{\infty} \frac{\sqrt{n^3 + 1}}{n^2 + 4}$ converge?
Show answer
Leading terms: $\sqrt{n^3}/n^2 = n^{3/2}/n^2 = 1/n^{1/2}$. Compute $\lim \dfrac{\sqrt{n^3+1}/(n^2+4)}{1/\sqrt{n}} = \lim \dfrac{\sqrt{n^4+n}}{n^2+4} = 1$. Since $\sum 1/\sqrt{n}$ diverges ($p=1/2$), the series diverges.