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Limit Comparison Test: When lim(aₙ/bₙ) is Finite > 0

Limit Comparison Test

A convergence test often used when the terms of a series are rational functions. Essentially, the test determines whether a series is "about as good" as a "good" series or "about as bad" as a "bad" series. The "good" or "bad" series is often a p-series.

See also

Comparison test, convergent series, divergent series, limit, infinity, positive series

Key Formula

Let L=lim⁡n→∞anbnIf 0<L<∞, then ∑an and ∑bn both converge or both diverge.\begin{gathered}\text{Let } L = \lim_{n \to \infty} \frac{a_n}{b_n}\\\text{If } 0 < L < \infty, \text{ then } \sum a_n \text{ and } \sum b_n \text{ both converge or both diverge.}\end{gathered}
Where:
  • ana_n = The general term of the series you are testing (must be positive for all sufficiently large n)
  • bnb_n = The general term of a known comparison series (must be positive for all sufficiently large n)
  • LL = The limit of the ratio a_n / b_n as n approaches infinity

Worked Example

Problem: Determine whether the series converges or diverges: ∑ (3n² + 5) / (n⁴ + 2n + 1) from n = 1 to infinity.
Step 1: Identify the dominant behavior. For large n, the numerator behaves like 3n² and the denominator behaves like n⁴, so the terms behave like 3n²/n⁴ = 3/n². Choose the comparison series b_n = 1/n², which is a convergent p-series (p = 2 > 1).
an=3n2+5n4+2n+1,bn=1n2a_n = \frac{3n^2 + 5}{n^4 + 2n + 1}, \quad b_n = \frac{1}{n^2}
Step 2: Compute the limit of the ratio a_n / b_n as n approaches infinity.
L=lim⁡n→∞anbn=lim⁡n→∞3n2+5n4+2n+1⋅n2=lim⁡n→∞3n4+5n2n4+2n+1L = \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{3n^2 + 5}{n^4 + 2n + 1} \cdot n^2 = \lim_{n \to \infty} \frac{3n^4 + 5n^2}{n^4 + 2n + 1}
Step 3: Divide the numerator and denominator by n⁴ to evaluate the limit.
L=lim⁡n→∞3+5n21+2n3+1n4=31=3L = \lim_{n \to \infty} \frac{3 + \frac{5}{n^2}}{1 + \frac{2}{n^3} + \frac{1}{n^4}} = \frac{3}{1} = 3
Step 4: Since L = 3, which satisfies 0 < L < ∞, the Limit Comparison Test tells us that ∑ a_n and ∑ b_n share the same convergence behavior. Because ∑ 1/n² converges, the original series also converges.
Answer: The series ∑ (3n² + 5) / (n⁴ + 2n + 1) converges by the Limit Comparison Test with ∑ 1/n².

Another Example

Problem: Determine whether the series converges or diverges: ∑ 1 / (2n − 1) from n = 1 to infinity.
Step 1: For large n, the term 1/(2n − 1) behaves like 1/(2n). The harmonic series ∑ 1/n diverges, so choose b_n = 1/n as the comparison series.
an=12n−1,bn=1na_n = \frac{1}{2n - 1}, \quad b_n = \frac{1}{n}
Step 2: Compute the limit of the ratio.
L=lim⁡n→∞1/(2n−1)1/n=lim⁡n→∞n2n−1=lim⁡n→∞12−1/n=12L = \lim_{n \to \infty} \frac{1/(2n-1)}{1/n} = \lim_{n \to \infty} \frac{n}{2n - 1} = \lim_{n \to \infty} \frac{1}{2 - 1/n} = \frac{1}{2}
Step 3: Since L = 1/2, which is finite and positive, both series share the same behavior. Because ∑ 1/n diverges, the original series also diverges.
Answer: The series ∑ 1/(2n − 1) diverges by the Limit Comparison Test with the harmonic series ∑ 1/n.

Frequently Asked Questions

What happens if the limit L equals 0 or infinity in the Limit Comparison Test?
If L = 0 and ∑ b_n converges, then ∑ a_n also converges. If L = ∞ and ∑ b_n diverges, then ∑ a_n also diverges. In other cases (L = 0 with ∑ b_n divergent, or L = ∞ with ∑ b_n convergent), the test is inconclusive and you need a different method.
How do I choose the comparison series b_n?
Look at the dominant terms in a_n for large n. Keep only the highest powers in the numerator and denominator, then simplify. The resulting expression is your b_n. For rational functions, this almost always yields a p-series of the form 1/n^p, whose convergence behavior you can determine immediately.

Limit Comparison Test vs. Direct Comparison Test

The Direct Comparison Test requires you to establish an inequality a_n ≤ b_n (for convergence) or a_n ≥ b_n (for divergence) for all sufficiently large n. The Limit Comparison Test only requires computing a limit, which is often easier when the algebra of bounding terms is messy. However, the Direct Comparison Test can sometimes work in cases where the limit does not exist.

Why It Matters

The Limit Comparison Test is one of the most frequently used convergence tests in calculus because it handles rational-type expressions with ease. Instead of wrestling with inequalities to bound terms above or below, you simply take a limit—a skill you've practiced extensively. It is especially powerful for series whose terms are quotients of polynomials, roots of polynomials, or other expressions with clear dominant behavior at infinity.

Common Mistakes

Mistake: Applying the test when a_n or b_n takes negative values.
Correction: The Limit Comparison Test requires both a_n > 0 and b_n > 0 for all sufficiently large n. If terms alternate in sign, use the Alternating Series Test or absolute convergence instead.
Mistake: Getting L = 0 or L = ∞ and concluding both series behave the same.
Correction: The conclusion that both series share the same behavior only applies when 0 < L < ∞. When L = 0 or L = ∞, only one direction of the conclusion holds, and in some cases the test is inconclusive.

Related Terms