Calculus II · Unit 4A · lesson

The Limit Comparison Test

Identify dominant term scale, choose an effective benchmark, and use a positive finite ratio limit to transfer convergence behavior.

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Section overview

Positive-series tests and error estimates

What this section is building

Identify dominant term scale, choose an effective benchmark, and use a positive finite ratio limit to transfer convergence behavior.

Notice

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

Decide

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

Avoid

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

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 $n$ , 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 $p$ -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 $b_n$ from a familiar family with that scale. Then compute $\lim a_n/b_n$ ; 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 $a_n/b_n\to L$ with $0<L<\infty$ , 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 $n$ ; 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, $\varepsilon=L/2$ . Eventually

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

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

Concept

Limit Comparison Test

For positive $a_n,b_n$ , if

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

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

Guided walkthrough

Dominant powers

Let

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

Then

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

Since $\sum1/n^2$ converges, so does $\sum a_n$ .

Worked example

Leading powers select the benchmark

Test

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

The dominant behavior is $4n^2/n^4=4/n^2$ , so use $b_n=1/n^2$ . Then

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

Because $4$ is positive and finite and $\sum1/n^2$ converges, the given series converges.

Common mistake

State the benchmark series, not only the ratio

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

Interactive checku4a-limit_comparison_test-01

For $a_n=(3n+1)/(n^3-4)$ and $b_n=1/n^2$ , find $\lim a_n/b_n$ .

Your work stays on this device. No account or AI grader is used.

Show hint

Multiply $a_n$ by $n^2$ and compare highest powers.

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

Exercise

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

Exercise

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

Exercise

Explain why a ratio limit of $1$ is called asymptotic equivalence.

Exercise

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

Source & rights

Original instruction with traceable references.

BetterGrades-original; no direct adaptation declared in the verified handoff.

Reference textbooks remain rights-separated and are not published as application assets. Any direct adaptation requires separate identification and attribution.

The Limit Comparison Test

Explanation

Asymptotic sameness is often enough

Decision

Choose the benchmark before taking the limit

Bridge

Asymptotic proportionality replaces a hard inequality

Proof idea

A limit near L produces two ordinary comparisons

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

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

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

Concept

Limit Comparison Test

For positive $a_n,b_n$ , if

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

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

Guided walkthrough

Dominant powers

Let

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

Then

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

Since $\sum1/n^2$ converges, so does $\sum a_n$ .

Worked example

Leading powers select the benchmark

Test

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

The dominant behavior is $4n^2/n^4=4/n^2$ , so use $b_n=1/n^2$ . Then

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

Because $4$ is positive and finite and $\sum1/n^2$ converges, the given series converges.

Common mistake

State the benchmark series, not only the ratio

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

Interactive checku4a-limit_comparison_test-01

For $a_n=(3n+1)/(n^3-4)$ and $b_n=1/n^2$ , find $\lim a_n/b_n$ .

Your work stays on this device. No account or AI grader is used.

Show hint

Multiply $a_n$ by $n^2$ and compare highest powers.

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

Exercise

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

Exercise

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

Exercise

Explain why a ratio limit of $1$ is called asymptotic equivalence.

Exercise

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

Source & rights

Original instruction with traceable references.

BetterGrades-original; no direct adaptation declared in the verified handoff.

Reference textbooks remain rights-separated and are not published as application assets. Any direct adaptation requires separate identification and attribution.

The Limit Comparison Test

Related resources

Original instruction with traceable references.

The Limit Comparison Test

Related resources

Original instruction with traceable references.