Calculus II · Unit 4A · lesson

The Limit Comparison Test

Concept

Learning objectives

use asymptotic ratios to compare positive series when direct inequalities are awkward.

The Limit Comparison Test

Explanation

Asymptotic sameness is often enough

Two positive sequences can differ in lower-order details while having the same long-run scale. If their ratio approaches a finite positive constant, then neither is eventually dramatically larger than the other. Their series therefore share the same convergence behavior.

Limit comparison is particularly efficient for rational functions of nn, radicals, and expressions whose dominant terms are obvious. It classifies convergence but usually does not produce the sum. The comparison sequence should be simple and known, most often a pp-series or geometric series. A ratio limit of zero or infinity may still provide information, but the standard equivalence conclusion requires a positive finite constant.

Decision

Choose the benchmark before taking the limit

Ignore lower-order terms and identify the dominant power, logarithm, or exponential behavior. Select bnb_n from a familiar family with that scale. Then compute liman/bn\lim a_n/b_n; a positive finite limit means the two series share a convergence type.

Bridge

Asymptotic proportionality replaces a hard inequality

Exact term-by-term inequalities can be awkward when formulas contain several competing pieces. If an/bnLa_n/b_n\to L with 0<L<0<L<\infty, then the terms are eventually comparable by positive constant multiples. Neither series can have a fundamentally different convergence behavior from the other.

The benchmark should match the dominant size of the target term. For rational powers, compare leading powers of nn; for radicals, factor the highest power inside the root. The limit must be positive and finite. A limit of zero or infinity may still suggest a one-sided comparison, but it is not the standard Limit Comparison Test conclusion.

Proof idea

A limit near L produces two ordinary comparisons

Choose, for example, ε=L/2\varepsilon=L/2. Eventually

L2<anbn<3L2,\frac L2<\frac{a_n}{b_n}<\frac{3L}{2},

so (L/2)bn<an<(3L/2)bn(L/2)b_n<a_n<(3L/2)b_n. Direct comparison then transfers convergence or divergence.

Concept

Limit Comparison Test

For positive an,bna_n,b_n, if

limnanbn=cwith 0<c<,\lim_{n\to\infty}\frac{a_n}{b_n}=c \quad\text{with }0<c<\infty,

then an\sum a_n and bn\sum b_n either both converge or both diverge.

Guided walkthrough

Dominant powers

Let

an=3n+1n34,bn=1n2.a_n=\frac{3n+1}{n^3-4},\qquad b_n=\frac1{n^2}.

Then

anbn=n2(3n+1)n343.\frac{a_n}{b_n}=\frac{n^2(3n+1)}{n^3-4}\to3.

Since 1/n2\sum1/n^2 converges, so does an\sum a_n.

Worked example

Leading powers select the benchmark

Test

n=14n2+1n4+7.\sum_{n=1}^{\infty}\frac{4n^2+1}{n^4+7}.

The dominant behavior is 4n2/n4=4/n24n^2/n^4=4/n^2, so use bn=1/n2b_n=1/n^2. Then

limnanbn=limnn2(4n2+1)n4+7=4.\lim_{n\to\infty}\frac{a_n}{b_n} =\lim_{n\to\infty}\frac{n^2(4n^2+1)}{n^4+7}=4.

Because 44 is positive and finite and 1/n2\sum1/n^2 converges, the given series converges.

Common mistake

State the benchmark series, not only the ratio

A correct solution identifies bnb_n, evaluates the ratio limit, states that the limit is positive and finite, and classifies bn\sum b_n.

Interactive checku4a-limit_comparison_test-01

For an=(3n+1)/(n34)a_n=(3n+1)/(n^3-4) and bn=1/n2b_n=1/n^2, find liman/bn\lim a_n/b_n.

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Show hint

Multiply ana_n by n2n^2 and compare highest powers.

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Exercise

Classify (n+2)/(n2+1)\sum(n+2)/(n^2+1).

Exercise

Compare 1/n2+n1/\sqrt{n^2+n} with 1/n1/n.

Exercise

Explain why a ratio limit of 11 is called asymptotic equivalence.

Exercise

Give an example where direct comparison is easier than limit comparison.

After the explanation

Use the section idea

Reading lens

Compare positive terms by long-run size and select a benchmark whose convergence behavior is already known.

Mental model

Direct comparison transfers inequalities; limit comparison transfers asymptotic scale; the integral test links sums to accumulated area.

Decision

Use a clean inequality when available, asymptotic comparison when ratios stabilize, and the integral test when a matching decreasing function is natural.

Common trap

Reversing the direction needed to prove convergence or divergence, or forgetting an integral-test remainder condition.

Check yourself

Does your benchmark support the conclusion in the direction you claim, and are all hypotheses stated?

Source & rights

Original instruction with traceable references.

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