Questions: Limit Comparison Test

Question 1

To test convergence of Σ (3n² + 7)/(n⁴ − n + 2), a student chooses bₙ = 1/n² and computes lim(aₙ/bₙ) = 3. What conclusion follows?

• A. The test is inconclusive — the limit must equal exactly 1 for the Limit Comparison Test to apply
• B. The series diverges because the limit 3 is greater than 1
• C. Since lim(aₙ/bₙ) = 3 (positive and finite) and Σ 1/n² converges (p-series, p = 2), the series Σaₙ also converges
• D. The series converges, but only after dividing by 3 to normalize the limit to 1

Answer: (C)

The Limit Comparison Test requires lim(aₙ/bₙ) = c where 0 < c < ∞ — any positive finite value works, not only c = 1. If the limit is 3, then aₙ ≈ 3bₙ for large n, meaning the two series are proportional and must behave identically (both converge or both diverge). Since Σ1/n² converges (p = 2 > 1), so does Σaₙ. Option A is the most common misconception about the test.

Question 2

A student applies the Limit Comparison Test to a series Σaₙ using comparison series Σbₙ and finds lim(aₙ/bₙ) = 0. What does this tell them?

• A. Σaₙ converges, because its terms are smaller than bₙ in the limit
• B. Σaₙ diverges, because a limit of 0 indicates the ratio collapses
• C. The test is inconclusive for this choice of bₙ — the student should try a comparison series that grows more slowly (matches aₙ's rate better)
• D. Σaₙ and Σbₙ converge and diverge oppositely

Answer: (C)

When lim(aₙ/bₙ) = 0, the terms aₙ grow much more slowly than bₙ — the comparison series is too large. The Limit Comparison Test only gives a conclusion when the limit is a positive finite number. A limit of 0 or ∞ is inconclusive: you cannot infer convergence or divergence directly. The student should choose a smaller bₙ that more closely matches the growth rate of aₙ. Option A is tempting but wrong — even if aₙ < bₙ, a smaller divergent series can still diverge (e.g., aₙ = 1/n and bₙ = 1).

Question 3

The Limit Comparison Test is more flexible than the Direct Comparison Test because it only requires the ratio aₙ/bₙ to approach a positive finite constant, rather than a termwise inequality aₙ ≤ bₙ.

• T. True
• F. False

Answer: True

True. The Direct Comparison Test needs aₙ ≤ bₙ (or aₙ ≥ bₙ) for all n — a pointwise condition that can be difficult or impossible to establish, especially when the numerator or denominator involves differences. The Limit Comparison Test only needs eventual proportionality: if aₙ/bₙ → c > 0, then for large n, aₙ ≈ c·bₙ, and the two series necessarily share convergence behavior. This avoids the need for any termwise inequality.

Question 4

If lim(aₙ/bₙ) = 0 and Σbₙ diverges, then Σaₙ is expected to also diverge.

• T. True
• F. False

Answer: False

False. A limit of 0 means aₙ grows much slower than bₙ — the series Σaₙ could converge even though Σbₙ diverges. For example, aₙ = 1/n² and bₙ = 1/n: lim(aₙ/bₙ) = lim(n/n²) = lim(1/n) = 0, Σ1/n diverges, but Σ1/n² converges. The Limit Comparison Test only provides a conclusion when the limit is strictly between 0 and ∞. At the boundary values 0 and ∞, no conclusion can be drawn about the original series from the chosen comparison alone.

Question 5

Explain why the Limit Comparison Test fails to give a conclusion when lim(aₙ/bₙ) = ∞, and describe what this tells you about your choice of comparison series bₙ.

Model answer: When lim(aₙ/bₙ) = ∞, the terms aₙ grow much faster than bₙ — the comparison series bₙ is too small. The test cannot conclude convergence or divergence because: a series growing faster than a convergent series could still converge (e.g., faster than 1/n³ but still summable), and a series growing faster than a divergent series clearly diverges — but we don't know which case we're in without more information. The fix is to choose a larger bₙ that grows at the same asymptotic rate as aₙ (by identifying the dominant terms in the numerator and denominator of aₙ). The goal is a bₙ such that lim(aₙ/bₙ) = c for some 0 < c < ∞.

The skill in applying the test is matching growth rates. For rational sequences, extract the leading-term ratio: if aₙ = (5n³ + 2n)/(n⁵ + 7), the dominant terms give 5n³/n⁵ = 5/n², so try bₙ = 1/n². The constant 5 doesn't affect the convergence conclusion. A limit of ∞ with bₙ = 1/n² means you need a larger comparison, perhaps 1/n. A limit of 0 means you need a smaller comparison, perhaps 1/n³.