Dove's LibraryDove's Library

10.6 - Comparison Tests for Convergence

Introduction

Comparison tests let you figure out whether a complicated series converges by comparing it to a simpler series whose behavior you already know.
Use them when: - The series looks nasty to integrate - It isn’t geometric or a p-series - You want to avoid ugly integration
Basic strategy: compare the unknown series to a known benchmark, usually a geometric series or a p-series.

The Direct Comparison Test (DCT)

Conditions: let ∑an\sum a_n and ∑bn\sum b_n be series with positive terms, where 0<an≤bn0 < a_n \leq b_n for all n≥kn \geq k.
Convergence rule: if the larger series converges, the smaller one does too.
If ∑bn converges, then ∑an converges\text{If } \sum b_n \text{ converges, then } \sum a_n \text{ converges}
Divergence rule: if the smaller series diverges, the larger one does too.
If ∑an diverges, then ∑bn diverges\text{If } \sum a_n \text{ diverges, then } \sum b_n \text{ diverges}

Visual Understanding

10.6 - Comparison Tests for Convergence
Intuition for convergence: if the bigger series adds up to something finite, the smaller series (squeezed underneath it) has to be finite too.
10.6 - Comparison Tests for Convergence
Intuition for divergence: if the smaller series blows up, the bigger series (which sits above it) has to blow up as well.

What to Compare To

Key principle: simplify by keeping only the dominant terms.
For large nn: - Lower-order terms become negligible - Constants don’t meaningfully affect the growth rate - Focus on the highest power of nn in the numerator and denominator
Example: for ∑3n2+5n+17n4+2n2+8\sum \frac{3n^2 + 5n + 1}{7n^4 + 2n^2 + 8}, the dominant terms give:
3n27n4=37n2\frac{3n^2}{7n^4} = \frac{3}{7n^2}
Drop the constant and compare to 1n2\frac{1}{n^2}, a p-series with p=2>1p = 2 > 1, which converges.

Examples: Direct Comparison Test

Example 1: exponential comparison
Does ∑n=1∞15+3n\sum_{n=1}^{\infty} \frac{1}{5 + 3^n} converge or diverge?
Step 1: find a comparison series. For n≥1n \geq 1, 5+3n>3n5 + 3^n > 3^n, so:
15+3n<13n\frac{1}{5 + 3^n} < \frac{1}{3^n}
Step 2: analyze the comparison series
∑n=1∞13n=∑n=1∞(13)n\sum_{n=1}^{\infty} \frac{1}{3^n} = \sum_{n=1}^{\infty} \left(\frac{1}{3}\right)^n
Geometric with r=13r = \frac{1}{3}, and since ∣r∣<1|r| < 1, it converges.
Step 3: apply DCT. We have:
0<15+3n<13n0 < \frac{1}{5 + 3^n} < \frac{1}{3^n}
The larger series converges, so the smaller one converges by DCT.
Example 2: careful with the direction of the inequality
Does ∑n=1∞12n−3\sum_{n=1}^{\infty} \frac{1}{2^n - 3} converge or diverge?
Attempt 1: compare to 12n\frac{1}{2^n}. For n≥2n \geq 2, 2n−3<2n2^n - 3 < 2^n, so:
12n−3>12n\frac{1}{2^n - 3} > \frac{1}{2^n}
We know ∑12n\sum \frac{1}{2^n} converges, but our series is bigger than it. That tells us nothing. A smaller series converging says nothing about a larger one.
Attempt 2: find an upper bound instead. For n≥3n \geq 3:
2n−3>2n2=2n−12^n - 3 > \frac{2^n}{2} = 2^{n-1}
so:
12n−3<12n−1=22n\frac{1}{2^n - 3} < \frac{1}{2^{n-1}} = \frac{2}{2^n}
Now ∑22n=2∑12n\sum \frac{2}{2^n} = 2\sum \frac{1}{2^n} converges (geometric). Since:
0<12n−3<22n0 < \frac{1}{2^n - 3} < \frac{2}{2^n}
the series ∑12n−3\sum \frac{1}{2^n - 3} converges by DCT.
What about the first two terms? At n=1n = 1 the term is 12−3=−1\frac{1}{2 - 3} = -1, which is negative, and the inequality only starts at n=3n = 3. That’s fine. DCT only needs 0<an≤bn0 < a_n \leq b_n from some point on, and the first two terms add a finite amount (−1+1=0-1 + 1 = 0) that can’t change whether the series converges.
Example 3: using a p-series comparison
Does ∑n=1∞n+23n3+5\sum_{n=1}^{\infty} \frac{n + 2}{3n^3 + 5} converge or diverge?
Step 1: identify the dominant behavior. For large nn:
n+23n3+5≈n3n3=13n2\frac{n + 2}{3n^3 + 5} \approx \frac{n}{3n^3} = \frac{1}{3n^2}
Step 2: establish the inequality. For n≥1n \geq 1: - n+2≤3nn + 2 \leq 3n - 3n3+5≥3n33n^3 + 5 \geq 3n^3
So:
n+23n3+5≤3n3n3=1n2\frac{n + 2}{3n^3 + 5} \leq \frac{3n}{3n^3} = \frac{1}{n^2}
Step 3: apply DCT. ∑1n2\sum \frac{1}{n^2} is a p-series with p=2>1p = 2 > 1, so it converges. Since:
0<n+23n3+5≤1n20 < \frac{n + 2}{3n^3 + 5} \leq \frac{1}{n^2}
the series converges by DCT.

The Limit Comparison Test (LCT)

The LCT is usually easier, because you don’t have to figure out which series is bigger.
Let ∑an\sum a_n and ∑bn\sum b_n be series with positive terms. If:
lim⁡n→∞anbn=L\lim_{n \to \infty} \frac{a_n}{b_n} = L
where LL is finite and positive (0<L<∞0 < L < \infty), then:
Both series converge, or both diverge\boxed{\text{Both series converge, or both diverge}}
The big advantage: no inequality to establish.

Examples: Limit Comparison Test

Example 4: polynomial vs. p-series
Does ∑n=1∞4n2−79n4+2n+1\sum_{n=1}^{\infty} \frac{4n^2 - 7}{9n^4 + 2n + 1} converge or diverge?
Step 1: choose a comparison series. Keep the dominant terms:
4n29n4=49n2≈1n2\frac{4n^2}{9n^4} = \frac{4}{9n^2} \approx \frac{1}{n^2}
Compare to bn=1n2b_n = \frac{1}{n^2}, a p-series with p=2>1p = 2 > 1, which converges.
Step 2: calculate the limit
L=lim⁡n→∞4n2−79n4+2n+11n2=lim⁡n→∞(4n2−7)⋅n29n4+2n+1=lim⁡n→∞4n4−7n29n4+2n+1L = \lim_{n \to \infty} \frac{\frac{4n^2 - 7}{9n^4 + 2n + 1}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{(4n^2 - 7) \cdot n^2}{9n^4 + 2n + 1} = \lim_{n \to \infty} \frac{4n^4 - 7n^2}{9n^4 + 2n + 1}
Divide top and bottom by n4n^4:
=lim⁡n→∞4−7n29+2n3+1n4=49= \lim_{n \to \infty} \frac{4 - \frac{7}{n^2}}{9 + \frac{2}{n^3} + \frac{1}{n^4}} = \frac{4}{9}
Step 3: apply LCT. Since L=49L = \frac{4}{9} is finite and positive, and ∑1n2\sum \frac{1}{n^2} converges, the original series converges by LCT.
Example 5: radical expression
Does ∑n=1∞2n+37n3+5\sum_{n=1}^{\infty} \frac{2n + 3}{\sqrt{7n^3 + 5}} converge or diverge?
Step 1: choose a comparison series
2n7n3=2n7⋅n3/2=27⋅n−1/2≈1n1/2\frac{2n}{\sqrt{7n^3}} = \frac{2n}{\sqrt{7} \cdot n^{3/2}} = \frac{2}{\sqrt{7}} \cdot n^{-1/2} \approx \frac{1}{n^{1/2}}
Compare to bn=1nb_n = \frac{1}{\sqrt{n}}, a p-series with p=12<1p = \frac{1}{2} < 1, which diverges.
Step 2: calculate the limit
L=lim⁡n→∞2n+37n3+51n=lim⁡n→∞(2n+3)n7n3+5L = \lim_{n \to \infty} \frac{\frac{2n + 3}{\sqrt{7n^3 + 5}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{(2n + 3)\sqrt{n}}{\sqrt{7n^3 + 5}}=lim⁡n→∞n(2n+3)n37+5n3=lim⁡n→∞2n+3n7+5n3=lim⁡n→∞2+3n7+5n3=27= \lim_{n \to \infty} \frac{\sqrt{n}(2n + 3)}{\sqrt{n^3}\sqrt{7 + \frac{5}{n^3}}} = \lim_{n \to \infty} \frac{2n + 3}{n\sqrt{7 + \frac{5}{n^3}}} = \lim_{n \to \infty} \frac{2 + \frac{3}{n}}{\sqrt{7 + \frac{5}{n^3}}} = \frac{2}{\sqrt{7}}
Step 3: apply LCT. L=27L = \frac{2}{\sqrt{7}} is finite and positive, and ∑1n\sum \frac{1}{\sqrt{n}} diverges, so the original series diverges by LCT.
Example 6: exponential vs. polynomial
Does ∑n=1∞5n4n+n2\sum_{n=1}^{\infty} \frac{5n}{4^n + n^2} converge or diverge?
Step 1: choose a comparison. Exponentials dominate polynomials, so for large nn, 4n≫n24^n \gg n^2:
5n4n+n2≈5n4n\frac{5n}{4^n + n^2} \approx \frac{5n}{4^n}
Compare to bn=n4nb_n = \frac{n}{4^n}.
Step 2: confirm the comparison series converges. Note that n<2nn < 2^n for all n≥1n \geq 1, so:
n4n<2n4n=12n\frac{n}{4^n} < \frac{2^n}{4^n} = \frac{1}{2^n}
And ∑12n\sum \frac{1}{2^n} is geometric with r=12r = \frac{1}{2}, which converges. So ∑n4n\sum \frac{n}{4^n} converges by DCT. (You can also confirm this with the ratio test, coming up in 10.8.)
Step 3: calculate the limit
L=lim⁡n→∞5n4n+n2n4n=lim⁡n→∞5n⋅4nn(4n+n2)=lim⁡n→∞5⋅4n4n+n2L = \lim_{n \to \infty} \frac{\frac{5n}{4^n + n^2}}{\frac{n}{4^n}} = \lim_{n \to \infty} \frac{5n \cdot 4^n}{n(4^n + n^2)} = \lim_{n \to \infty} \frac{5 \cdot 4^n}{4^n + n^2}
Divide by 4n4^n:
=lim⁡n→∞51+n24n=51+0=5= \lim_{n \to \infty} \frac{5}{1 + \frac{n^2}{4^n}} = \frac{5}{1 + 0} = 5
Step 4: apply LCT. L=5L = 5 is finite and positive, and ∑n4n\sum \frac{n}{4^n} converges, so the original series converges by LCT.
Example 7: fractional powers
Does ∑n=1∞nn3+n\sum_{n=1}^{\infty} \frac{n}{\sqrt{n^3 + n}} converge or diverge?
Step 1: simplify and choose a comparison
nn3+n≈nn3=nn3/2=1n\frac{n}{\sqrt{n^3 + n}} \approx \frac{n}{\sqrt{n^3}} = \frac{n}{n^{3/2}} = \frac{1}{\sqrt{n}}
Compare to bn=1nb_n = \frac{1}{\sqrt{n}}, a p-series with p=12<1p = \frac{1}{2} < 1, which diverges.
Step 2: calculate the limit
L=lim⁡n→∞nn3+n1n=lim⁡n→∞nnn3+n=lim⁡n→∞n3/2n3(1+1n2)L = \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + n}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n\sqrt{n}}{\sqrt{n^3 + n}} = \lim_{n \to \infty} \frac{n^{3/2}}{\sqrt{n^3\left(1 + \frac{1}{n^2}\right)}}=lim⁡n→∞n3/2n3/21+1n2=lim⁡n→∞11+1n2=1= \lim_{n \to \infty} \frac{n^{3/2}}{n^{3/2}\sqrt{1 + \frac{1}{n^2}}} = \lim_{n \to \infty} \frac{1}{\sqrt{1 + \frac{1}{n^2}}} = 1
Step 3: apply LCT. L=1L = 1 is finite and positive, and ∑1n\sum \frac{1}{\sqrt{n}} diverges, so the original series diverges by LCT.

DCT vs. LCT

Feature
Direct Comparison Test
Limit Comparison Test
Setup
Need 0<an<bn0 < a_n < b_n (or the reverse)
Calculate lim⁡n→∞anbn\lim_{n \to \infty} \frac{a_n}{b_n}
Difficulty
Must establish an inequality
Just compute a limit
When the larger converges
The smaller converges too
Both behave the same
When the smaller diverges
The larger diverges too
Both behave the same
Advantage
Direct logical connection
Don’t need the inequality direction
Best for
Clear inequalities
Complicated expressions

Strategy Guide

Step 1: identify the dominant behavior. Keep the highest power in the numerator and denominator, and drop the coefficients.
Step 2: choose a benchmark series. - p-series: ∑1np\sum \frac{1}{n^p} - Geometric: ∑rn\sum r^n - Any other series you already know
Step 3: decide on a test. - DCT if the inequality is easy to establish - LCT if computing the limit is straightforward (usually the easier route)
Step 4: apply it. - DCT: prove the inequality, then conclude - LCT: compute the limit, check that 0<L<∞0 < L < \infty, then conclude

Common Benchmarks

Geometric series: ∑arn\sum ar^n converges if ∣r∣<1|r| < 1. Use for exponential terms like 2n2^n or e−ne^{-n}.
p-series: ∑1np\sum \frac{1}{n^p} converges if p>1p > 1, diverges if p≤1p \leq 1. Use for polynomial-like terms.
Harmonic series: ∑1n\sum \frac{1}{n} diverges. Use for anything roughly like 1n\frac{1}{n}.

Key Takeaways

  • Comparison tests: compare an unknown series to a known benchmark
  • DCT requires an inequality: you have to show an<bna_n < b_n or an>bna_n > b_n
  • DCT logic: larger converges → smaller converges; smaller diverges → larger diverges
  • LCT uses limits: if lim⁡anbn=L\lim \frac{a_n}{b_n} = L is finite and positive, both series behave the same
  • LCT advantage: you don’t need to know which series is bigger
  • Dominant terms: focus on the highest powers when picking a comparison
  • Drop constants: coefficients don’t affect convergence behavior for large nn
  • Common benchmarks: p-series and geometric series
  • p-series: converges if p>1p > 1, diverges if p≤1p \leq 1
  • Geometric: converges if ∣r∣<1|r| < 1, diverges if ∣r∣≥1|r| \geq 1
  • Both tests require positive terms throughout

AP® is a registered trademark of the College Board, which is not affiliated with Dove's Library and was not involved in the production of, and does not endorse, this website.

Providing free, high-quality AP notes to help students succeed, earn college credit, and reach their educational goals, ensuring everyone has access to the resources they need.

© 2026 Standard Learning Group, Inc., which operates Dove's Library. All rights reserved.