Limit Comparison Test Explained Simply

The limit comparison test tells you whether two series share the same convergence fate by examining the limit of their term ratio. If that limit is a positive finite number, both series converge or both diverge. Compare an unknown series to a known one (a p-series or a geometric series) and the limit does the work. You do not need an inequality, which is why the limit comparison test is often easier than the direct comparison test. Both tests appear in Stewart's Calculus, section 11.4 (9th edition, 2016, Cengage) and in Paul's Online Math Notes (Comparison Test/Limit Comparison Test, 2003-2025, Lamar University). This covers choosing b_n and applying each test correctly, including the cases where one fails and the other succeeds.

The direct comparison test has two directions. If 0 ≤ a_n ≤ b_n for all n and Σ b_n converges, then Σ a_n converges. If a_n ≥ b_n ≥ 0 and Σ b_n diverges, then Σ a_n diverges. The inequality must run the correct way. Reverse it and you get no information. The test is named simply the Comparison Test in both Stewart (section 11.4) and Paul's Online Math Notes. Compare a_n to a known series b_n, typically a p-series Σ 1/n^p or a geometric series Σ ar^(n-1). A p-series converges if p > 1 and diverges if p ≤ 1. A geometric series converges if |r| < 1 and diverges if |r| ≥ 1. These are your benchmarks. The two directions are: (1) a_n ≤ b_n when b_n converges, and (2) a_n ≥ b_n when b_n diverges. If you cannot find an inequality that holds, the direct comparison test is inconclusive; try the limit comparison test.

The limit comparison test states: if lim_{n→∞} a_n/b_n = c where 0 < c < ∞, then both series Σ a_n and Σ b_n either converge or diverge together. The number c can be anything positive and finite. Unlike the direct comparison test, you do not need an inequality. But what happens when the limit is 0 or ∞? If the limit is 0 and Σ b_n converges, then Σ a_n also converges. If the limit is ∞ and Σ b_n diverges, then Σ a_n also diverges. These are less common but still useful. The core case is c being a positive finite number. Paul's Online Math Notes calls this test the Limit Comparison Test; Stewart uses the same name in section 11.4. The test works best when a_n is a rational function and you compare it to a p-series b_n. For example, a_n = (2n^2 + 3)/(n^3 + 1) behaves like 2/n, so compare to b_n = 1/n (a divergent p-series with p=1). The limit of a_n/b_n is 2, positive and finite, so Σ a_n diverges.

Choosing b_n is the step where most errors occur. The rule: keep only the dominant terms in the numerator and denominator of a_n. Drop lower-order terms. For a rational function like (3n^2 + 2n + 1)/(5n^3 - n + 7), the dominant terms are 3n^2 in the numerator and 5n^3 in the denominator, so b_n = n^2/n^3 = 1/n. That is a p-series with p=1, which diverges. For a_n = (n + 1)/(n^2 + 2n + 3), dominant term in numerator is n, denominator is n^2, so b_n = 1/n, p=1, diverges. For a_n = (n^2)/(n^4 + 1), b_n = 1/n^2, p=2, converges. The method also works when terms involve logarithms or roots: a_n = (sqrt(n) + 1)/(n + 2) becomes b_n = sqrt(n)/n = 1/n^(1/2), p=1/2, diverges. This is the dominant-term guide. If the limit of a_n/b_n is a positive finite number, you have chosen well. If the limit is 0 or ∞, you may need a different b_n. Use the limit comparison test calculator carefully: it gives the same c value, but you must interpret it correctly.

Direct Comparison Test

Determine convergence of Σ a_n where a_n = 1/(n^2 + 5). Compare to b_n = 1/n^2 (p=2, converges). For all n, 1/(n^2 + 5) ≤ 1/n^2. Since Σ b_n converges, Σ a_n converges.

Direct Comparison Test (Divergence Direction)

Determine convergence of Σ a_n where a_n = 1/(sqrt(n) - 1). For n ≥ 4, sqrt(n) - 1 ≤ sqrt(n), so 1/(sqrt(n) - 1) ≥ 1/sqrt(n) = 1/n^(1/2). b_n = 1/n^(1/2) is a p-series with p=1/2, diverges. Since a_n ≥ b_n and Σ b_n diverges, Σ a_n diverges.

Limit Comparison Test

Determine convergence of Σ a_n where a_n = (2n^2 + 3n)/(n^3 + 5n + 2). Dominant terms: 2n^2 and n^3, so b_n = 2/n. Compute limit of a_n/b_n = limit of (2n^2 + 3n)/(n^3 + 5n + 2) * (n/2) = limit of (2n^3 + 3n^2)/(2n^3 + 10n^2 + 4n) = 1. Since c = 1 is positive finite, both series share fate. b_n = 2/n diverges (p=1), so Σ a_n diverges.

Limit Comparison Test With a Logarithm

Determine convergence of Σ a_n where a_n = ln(n)/(n^2). Compare to b_n = 1/n^(3/2) (p=3/2, converges). The limit of a_n/b_n = limit of (ln(n) * n^(3/2))/n^2 = limit of ln(n)/n^(1/2) = 0. So limit is 0. Since Σ b_n converges and limit is 0, Σ a_n converges. A logarithm grows slower than any positive power, so the test works.

Direct comparison fails when you cannot find a simple inequality that holds in the correct direction. A common case: a_n = (sin(n) + 2)/(n^2). The sine term oscillates between -1 and 1, so a_n is between 1/n^2 and 3/n^2. You could bound a_n above by 3/n^2, but that only gives convergence if 3/n^2 converges, which it does. However, if a_n = (1 + cos(n))/(n), direct comparison is tricky because the numerator is between 0 and 2. You could bound above by 2/n (which diverges) giving no information, and below by 0 (which converges) also giving no information. The limit comparison test works here: compare a_n to 1/n. The limit of a_n/(1/n) = limit of (1 + cos(n)) = between 0 and 2, but not a fixed number because cos(n) oscillates. This is the failure case. For a_n = (1 + cos(pi*n))/(n), cos(pi*n) alternates between 1 and -1, so a_n is 0 every other term. The limit does not exist. The direct and limit comparison tests both require positive terms eventually; for alternating or oscillating terms, use the alternating series test or the ratio test instead. Another common failure: a_n = (n^2 + 1)/(n^3 + n). Direct comparison: compare to 1/n. For large n, a_n ≈ 1/n, but the inequality a_n ≤ 1/n fails because (n^2 + 1)/(n^3 + n) ≤ (n^2)/(n^3) = 1/n is true for all n≥1? Check: (n^2 + 1)/(n^3 + n) ≤ n^2/(n^3) = 1/n if and only if n(n^2 + 1) ≤ n^2(n^3 + n)? This is messy. The limit comparison test is simpler: compute limit of a_n/(1/n) = limit of (n^2 + 1)/(n^3 + n) * n = limit of (n^3 + n)/(n^3 + n) = 1. Since c=1 positive finite, Σ a_n diverges (because 1/n diverges). The direct comparison test is not wrong; it is just harder to apply. The limit comparison test saves time. This is why the limit comparison test is preferred on exams and in practice.