Mathwords logoMathwords

Comparison Test

Comparison Test

A convergence test which compares the series under consideration to a known series. Essentially, the test determines whether a series is "better" than a "good" series or "worse" than a "bad" series. The "good" or "bad" series is often a p-series.

 

If ∑ an , ∑ cn , and ∑ dn are all positive series, where ∑ cn converges and ∑ dn diverges, then:

1. If an ≤ cn for all n ≥ N for some fixed N, then ∑ an converges.

2. If an ≥ dn for all n ≥ N for some fixed N, then ∑ an diverges.

 

See also

Limit comparison test

Key Formula

If 0≤an≤cn for all n≥N and ∑cn converges, then ∑an converges.\text{If } 0 \le a_n \le c_n \text{ for all } n \ge N \text{ and } \sum c_n \text{ converges, then } \sum a_n \text{ converges.} If 0≤dn≤an for all n≥N and ∑dn diverges, then ∑an diverges.\text{If } 0 \le d_n \le a_n \text{ for all } n \ge N \text{ and } \sum d_n \text{ diverges, then } \sum a_n \text{ diverges.}
Where:
  • ana_n = The terms of the series you want to test
  • cnc_n = The terms of a known convergent series used for comparison
  • dnd_n = The terms of a known divergent series used for comparison
  • NN = A fixed positive integer beyond which the inequality holds

Worked Example

Problem: Determine whether the series ∑n=1∞1n2+3\sum_{n=1}^{\infty} \frac{1}{n^2 + 3} converges or diverges.
Step 1: Identify a known series to compare with. Since n2+3>n2n^2 + 3 > n^2 for all n≥1n \ge 1, we have:
1n2+3<1n2\frac{1}{n^2 + 3} < \frac{1}{n^2}
Step 2: Determine whether the comparison series converges. The series ∑1n2\sum \frac{1}{n^2} is a p-series with p=2>1p = 2 > 1, so it converges.
∑n=1∞1n2 converges\sum_{n=1}^{\infty} \frac{1}{n^2} \text{ converges}
Step 3: Apply the Comparison Test. We have 0≤an≤cn0 \le a_n \le c_n where an=1n2+3a_n = \frac{1}{n^2+3} and cn=1n2c_n = \frac{1}{n^2}, and the larger series converges. By the Comparison Test:
∑n=1∞1n2+3 converges\sum_{n=1}^{\infty} \frac{1}{n^2 + 3} \text{ converges}
Answer: The series ∑n=1∞1n2+3\sum_{n=1}^{\infty} \frac{1}{n^2 + 3} converges by the Comparison Test, since its terms are smaller than those of the convergent p-series ∑1n2\sum \frac{1}{n^2}.

Another Example

Problem: Determine whether the series ∑n=1∞1n−0.5\sum_{n=1}^{\infty} \frac{1}{\sqrt{n} - 0.5} converges or diverges.
Step 1: Look for a known divergent series to compare with. For n≥1n \ge 1, we have n−0.5<n\sqrt{n} - 0.5 < \sqrt{n}, which means:
1n−0.5>1n\frac{1}{\sqrt{n} - 0.5} > \frac{1}{\sqrt{n}}
Step 2: Check the comparison series. The series ∑1n=∑1n1/2\sum \frac{1}{\sqrt{n}} = \sum \frac{1}{n^{1/2}} is a p-series with p=12<1p = \frac{1}{2} < 1, so it diverges.
∑n=1∞1n diverges\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \text{ diverges}
Step 3: Apply the Comparison Test. Our series has terms larger than the divergent series ∑1n\sum \frac{1}{\sqrt{n}}, so by the Comparison Test:
∑n=1∞1n−0.5 diverges\sum_{n=1}^{\infty} \frac{1}{\sqrt{n} - 0.5} \text{ diverges}
Answer: The series ∑n=1∞1n−0.5\sum_{n=1}^{\infty} \frac{1}{\sqrt{n} - 0.5} diverges by the Comparison Test, since its terms are larger than those of the divergent p-series ∑1n\sum \frac{1}{\sqrt{n}}.

Frequently Asked Questions

When does the Comparison Test fail or give no conclusion?
The test is inconclusive in two situations: if your series is larger than a known convergent series, or if your series is smaller than a known divergent series. Being bigger than something convergent tells you nothing — you could still converge or diverge. Similarly, being smaller than something divergent gives no information. When this happens, try the Limit Comparison Test instead.
How do I choose the right series to compare with?
Focus on the dominant term in your expression for large nn. For instance, if an=1n3+5na_n = \frac{1}{n^3 + 5n}, the dominant term in the denominator is n3n^3, so compare with 1n3\frac{1}{n^3}. P-series (∑1np\sum \frac{1}{n^p}) and geometric series are the most common choices for comparison.

Direct Comparison Test vs. Limit Comparison Test

The Direct Comparison Test requires you to establish an inequality an≤cna_n \le c_n or an≥dna_n \ge d_n for all sufficiently large nn. The Limit Comparison Test instead computes lim⁡n→∞anbn\lim_{n \to \infty} \frac{a_n}{b_n}. If this limit is a finite positive number, both series share the same convergence behavior. The Limit Comparison Test is often easier to apply because you do not need to prove an inequality — you only need to evaluate a limit. However, if the limit equals 00 or ∞\infty, the Limit Comparison Test gives only a one-directional conclusion, similar to the direct test.

Why It Matters

The Comparison Test is one of the first and most intuitive convergence tests you learn in calculus. It builds directly on the idea that if a series is bounded above by something finite, it must also be finite — a principle that extends to many areas of analysis. Mastering it also trains you to identify the dominant behavior of a series for large nn, a skill that is essential for applying nearly every other convergence test.

Common Mistakes

Mistake: Comparing in the wrong direction — showing your series is smaller than a divergent series (or larger than a convergent series) and drawing a conclusion.
Correction: The test only works one way for each case. To prove convergence, your series must be smaller than a convergent series. To prove divergence, your series must be larger than a divergent series. The other directions are inconclusive.
Mistake: Applying the Comparison Test to series with negative terms.
Correction: The standard Comparison Test requires all terms to be non-negative (at least for all n≥Nn \ge N). For series with mixed-sign terms, consider using the Absolute Convergence Test or the Alternating Series Test instead.

Related Terms