Limit Comparison Test: Examples & Practice

The limit comparison test says that if $a_n \gt 0$ and $b_n \gt 0$ and $\lim_{n \to \infty} \dfrac{a_n}{b_n} = c$ , where c is a positive finite number, then $\sum a_n$ and $\sum b_n$ either both converge or both diverge. You pick $b_n$ as a simple series that behaves like $a_n$ for large n.

Before this lesson: Direct Comparison Test, p-Series and Limits at Infinity

Why the ratio of the terms matters

Take $a_n = \dfrac{1}{n^2 - 1}$ . For big n the −1 hardly matters, so $a_n$ is almost $\dfrac{1}{n^2}$ . The trouble is that $a_n$ is slightly bigger than $1/n^2$ , so the direct comparison test points the wrong way.

The limit comparison test avoids the inequality. It only asks whether the two sets of terms have a steady ratio in the long run. If $a_n$ is about c times $b_n$ for large n, then the two running totals grow or level off together.

The limit comparison test

\text{If } a_n, b_n \gt 0 \text{ and } \lim_{n \to \infty} \dfrac{a_n}{b_n} = c \text{ with } 0 \lt c \lt \infty, \text{ then } \sum a_n \text{ and } \sum b_n \text{ both converge or both diverge.}

Two edge cases. If the limit is 0 and $\sum b_n$ converges, then $\sum a_n$ converges (the terms $a_n$ are even smaller). If the limit is ∞ and $\sum b_n$ diverges, then $\sum a_n$ diverges. In the other combinations, a limit of 0 or ∞ tells you nothing, so choose a better $b_n$ .

Choosing b_n. Keep only the fastest-growing piece of the top and of the bottom. For a fraction of powers of n, $b_n = \dfrac{1}{n^p}$ , where p is the highest power on the bottom minus the highest power on the top.

How to use the limit comparison test

Check that the terms are positive, at least from some point on.
Pick $b_n$ from the fastest-growing pieces, and decide whether $\sum b_n$ converges (p-series or geometric series).
Find $\lim_{n \to \infty} \dfrac{a_n}{b_n}$ . To divide by $b_n$ , multiply by its reciprocal.
If the limit is a positive finite number, $\sum a_n$ does what $\sum b_n$ does.

Limit comparison test examples

Example 1: A term slightly bigger than 1/n²

Does $\sum_{n=2}^{\infty} \dfrac{1}{n^2 - 1}$ converge?

b_n = \dfrac{1}{n^2}

For large n, n² − 1 behaves like n². The terms are positive for n ≥ 2.

\dfrac{a_n}{b_n} = \dfrac{1}{n^2 - 1} \cdot \dfrac{n^2}{1} = \dfrac{n^2}{n^2 - 1}

Divide by multiplying by the reciprocal of bₙ.

\lim_{n \to \infty} \dfrac{n^2}{n^2 - 1} = \lim_{n \to \infty} \dfrac{1}{1 - \frac{1}{n^2}}

Divide the top and bottom by n².

\lim_{n \to \infty} \dfrac{1}{1 - \frac{1}{n^2}} = 1

1/n² goes to 0.

0 \lt 1 \lt \infty \quad \text{and} \quad \sum \dfrac{1}{n^2} \text{ converges}

The limit is positive and finite, and the comparison series is a p-series with p = 2.

Answer: the series converges by the limit comparison test. (It is also telescoping, with sum 3/4.)

Example 2: A rational term

Does $\sum_{n=1}^{\infty} \dfrac{3n + 5}{2n^3 - n + 4}$ converge?

b_n = \dfrac{n}{n^3} = \dfrac{1}{n^2}

Keep the fastest-growing pieces, 3n and 2n³, and drop the constants 3 and 2. The terms are positive for n ≥ 1.

\dfrac{a_n}{b_n} = \dfrac{3n + 5}{2n^3 - n + 4} \cdot n^2 = \dfrac{3n^3 + 5n^2}{2n^3 - n + 4}

Multiply by the reciprocal of bₙ (see multiplying polynomials).

\lim_{n \to \infty} \dfrac{3n^3 + 5n^2}{2n^3 - n + 4} = \lim_{n \to \infty} \dfrac{3 + \frac{5}{n}}{2 - \frac{1}{n^2} + \frac{4}{n^3}}

Divide the top and bottom by n³.

\lim_{n \to \infty} \dfrac{3 + \frac{5}{n}}{2 - \frac{1}{n^2} + \frac{4}{n^3}} = \dfrac{3}{2}

The small fractions go to 0.

\dfrac{3}{2} \gt 0 \text{ and } \sum \dfrac{1}{n^2} \text{ converges}

Positive, finite limit; p-series with p = 2.

Answer: the series converges. The ratio settles at 3/2, so for large n each term is about 1.5 times 1/n².

Dot plot of the ratio aₙ/bₙ = (3n³ + 5n²)/(2n³ − n + 4) for n = 1 to 30. The first value is 1.6, the ratio jumps to about 2.4 at n = 2, then drops and levels off just above a dashed line at 1.5. — The ratio of the terms to 1/n² settles at 3/2, a positive finite number, so both series behave the same way.

Example 3: A square root that acts like n

Does $\sum_{n=1}^{\infty} \dfrac{1}{\sqrt{n^2 + 3n}}$ converge?

b_n = \dfrac{1}{\sqrt{n^2}} = \dfrac{1}{n}

Inside the root, n² grows fastest.

\dfrac{a_n}{b_n} = \dfrac{n}{\sqrt{n^2 + 3n}}

Multiply by the reciprocal of bₙ.

\dfrac{n}{\sqrt{n^2 + 3n}} = \dfrac{1}{\sqrt{1 + \frac{3}{n}}}

Divide the top and bottom by n; inside the root, divide by n².

\lim_{n \to \infty} \dfrac{1}{\sqrt{1 + \frac{3}{n}}} = 1

3/n goes to 0.

\sum \dfrac{1}{n} \text{ diverges}

Harmonic series.

Answer: the series diverges by the limit comparison test.

Example 4: A trig term

Does $\sum_{n=1}^{\infty} \sin\left(\dfrac{1}{n}\right)$ converge?

\sin\left(\dfrac{1}{n}\right) \gt 0

For n ≥ 1, 1/n is between 0 and π, so the sine is positive.

b_n = \dfrac{1}{n}

For small angles, sin t is close to t.

\lim_{n \to \infty} \dfrac{\sin(1/n)}{1/n} = \lim_{t \to 0^+} \dfrac{\sin t}{t}

Let t = 1/n.

\lim_{t \to 0^+} \dfrac{\sin t}{t} = \lim_{t \to 0^+} \dfrac{\cos t}{1} = 1

Apply L’Hôpital’s rule.

\sum \dfrac{1}{n} \text{ diverges}

Harmonic series.

Answer: the series diverges, even though sin(1/n) goes to 0.

Example 5: Forest loss (word problem)

A model says a forest will lose $\dfrac{5n + 2}{n^2 + 1}$ hectares of trees in year n. If the model holds forever, is the total loss finite?

b_n = \dfrac{n}{n^2} = \dfrac{1}{n}

Keep the fastest-growing pieces, 5n and n², and drop the 5.

\dfrac{a_n}{b_n} = \dfrac{5n + 2}{n^2 + 1} \cdot n = \dfrac{5n^2 + 2n}{n^2 + 1}

Multiply by the reciprocal of bₙ.

\lim_{n \to \infty} \dfrac{5n^2 + 2n}{n^2 + 1} = \lim_{n \to \infty} \dfrac{5 + \frac{2}{n}}{1 + \frac{1}{n^2}} = 5

Divide the top and bottom by n².

\sum \dfrac{1}{n} \text{ diverges}

Harmonic series.

\sum_{n=1}^{\infty} \dfrac{5n + 2}{n^2 + 1} \text{ diverges}

The limit 5 is positive and finite.

Answer: no. The yearly loss shrinks (3.5 hectares in year 1, about 0.52 in year 10), but the total grows without bound because the losses shrink only like 5/n.

Common mistakes

Mistake	Example of the error	Fix
Concluding from a limit of 0 or ∞	Comparing $\frac{1}{n^2}$ with $\frac{1}{n}$ , getting 0, and saying “diverges”	A limit of 0 with a divergent $\sum b_n$ tells you nothing. Choose $b_n$ with the same growth, here $1/n^2$ .
Choosing b_n from the wrong powers	Using $b_n = \frac{1}{n}$ for $\frac{3n + 5}{2n^3 - n + 4}$	Subtract the top power from the bottom power: 3 − 1 = 2, so $b_n = 1/n^2$ .
Forgetting to say what Σb_n does	Finding the limit 3/2 and stopping	The test links the two series. State that $\sum 1/n^2$ converges, so the original series does too.
Using it on terms with mixed signs	Comparing $\frac{(-1)^n}{n}$ with $\frac{1}{n}$	Both sets of terms must be positive. Use the alternating series test instead.
Confusing it with the ratio test	Computing $\frac{a_{n+1}}{a_n}$	The limit comparison test divides by a different series, $\frac{a_n}{b_n}$ . The ratio test is a later lesson.
Worrying about which term goes on top	Thinking $\frac{b_n}{a_n}$ gives a different answer	Flipping the ratio turns c into 1/c. Both are positive and finite, so the conclusion is the same.

Related methods

Direct or limit comparison?

Try the direct comparison test when the inequality is easy, such as $\frac{1}{n^2 + 5} \lt \frac{1}{n^2}$ . Use the limit comparison test when the extra pieces push the inequality the wrong way, such as $\frac{1}{n^2 - 1}$ or $\frac{1}{2n - 1}$ .

Geometric benchmarks

For terms built from powers like $2^n$ and $3^n$ , compare with a geometric series. For example, $\frac{2^n + n}{3^n}$ behaves like $\left(\tfrac{2}{3}\right)^n$ , and the ratio of the two goes to 1.

Practice problems

Does $\sum_{n=1}^{\infty} \dfrac{n + 1}{n^3 + 2}$ converge?
Does $\sum_{n=1}^{\infty} \dfrac{1}{2n - 1}$ converge?
Does $\sum_{n=2}^{\infty} \dfrac{n^2}{n^4 - 3}$ converge?
Does $\sum_{n=1}^{\infty} \dfrac{1}{3^n - 2^n}$ converge?
Does $\sum_{n=1}^{\infty} \dfrac{\sqrt{n}}{n + 4}$ converge?
Does $\sum_{n=1}^{\infty} \dfrac{2n^2 + 1}{n^{5/2} + n}$ converge?
Does $\sum_{n=1}^{\infty} \tan\left(\dfrac{1}{n^2}\right)$ converge?
Does $\sum_{n=1}^{\infty} \left(1 - \cos\dfrac{1}{n}\right)$ converge?
Does $\sum_{n=1}^{\infty} \dfrac{1}{n^{1 + 1/n}}$ converge? (Hint: $n^{1 + 1/n} = n \cdot n^{1/n}$ and $n^{1/n} \to 1$ .)
A ball’s bounce on the nth try reaches $\dfrac{4n}{n^3 + 2n + 1}$ meters. Is the total of all the bounce heights finite?

Show answers

b_n = \dfrac{n}{n^3} = \dfrac{1}{n^2}

Keep the fastest-growing pieces.

\dfrac{a_n}{b_n} = \dfrac{(n + 1)n^2}{n^3 + 2} = \dfrac{n^3 + n^2}{n^3 + 2}

Multiply by n².

\lim_{n \to \infty} \dfrac{n^3 + n^2}{n^3 + 2} = 1

Divide the top and bottom by n³.

Answer: converges, since the limit is 1 and Σ1/n² converges

b_n = \dfrac{1}{n}

2n − 1 behaves like 2n; drop the constant 2.

\dfrac{a_n}{b_n} = \dfrac{n}{2n - 1}

Multiply by n.

\lim_{n \to \infty} \dfrac{n}{2n - 1} = \dfrac{1}{2}

Divide the top and bottom by n.

Answer: diverges, since the limit is 1/2 and the harmonic series diverges

b_n = \dfrac{n^2}{n^4} = \dfrac{1}{n^2}

Keep the fastest-growing pieces; the terms are positive for n ≥ 2.

\dfrac{a_n}{b_n} = \dfrac{n^4}{n^4 - 3}

Multiply by n².

\lim_{n \to \infty} \dfrac{n^4}{n^4 - 3} = 1

Divide the top and bottom by n⁴.

Answer: converges

b_n = \dfrac{1}{3^n}

3ⁿ grows much faster than 2ⁿ.

\dfrac{a_n}{b_n} = \dfrac{3^n}{3^n - 2^n} = \dfrac{1}{1 - \left(\tfrac{2}{3}\right)^n}

Multiply by 3ⁿ, then divide the top and bottom by 3ⁿ.

\lim_{n \to \infty} \dfrac{1}{1 - \left(\tfrac{2}{3}\right)^n} = 1

(2/3)ⁿ goes to 0.

Answer: converges, since Σ(1/3)ⁿ is a convergent geometric series

b_n = \dfrac{\sqrt{n}}{n} = \dfrac{1}{\sqrt{n}}

Keep the fastest-growing pieces.

\dfrac{a_n}{b_n} = \dfrac{\sqrt{n} \cdot \sqrt{n}}{n + 4} = \dfrac{n}{n + 4}

Multiply by √n.

\lim_{n \to \infty} \dfrac{n}{n + 4} = 1

Divide the top and bottom by n.

Answer: diverges, since Σ1/√n is a p-series with p = 1/2

b_n = \dfrac{n^2}{n^{5/2}} = \dfrac{1}{n^{1/2}}

Keep the fastest-growing pieces; 5/2 − 2 = 1/2.

\dfrac{a_n}{b_n} = \dfrac{(2n^2 + 1)n^{1/2}}{n^{5/2} + n} = \dfrac{2n^{5/2} + n^{1/2}}{n^{5/2} + n}

Multiply by n^(1/2).

\lim_{n \to \infty} \dfrac{2 + n^{-2}}{1 + n^{-3/2}} = 2

Divide the top and bottom by n^(5/2).

Answer: diverges, since Σ1/√n diverges

b_n = \dfrac{1}{n^2}

For small angles, tan t is close to t.

\lim_{t \to 0^+} \dfrac{\tan t}{t} = \lim_{t \to 0^+} \dfrac{\sec^2 t}{1} = 1

Let t = 1/n² and apply L’Hôpital’s rule.

Answer: converges, since Σ1/n² converges

b_n = \dfrac{1}{n^2}

For small t, 1 − cos t is close to t²/2.

\lim_{t \to 0^+} \dfrac{1 - \cos t}{t^2} = \lim_{t \to 0^+} \dfrac{\sin t}{2t}

Let t = 1/n; the form is 0/0, so apply L’Hôpital’s rule.

\lim_{t \to 0^+} \dfrac{\sin t}{2t} = \lim_{t \to 0^+} \dfrac{\cos t}{2} = \dfrac{1}{2}

Apply L’Hôpital’s rule again.

Answer: converges, since the limit is 1/2 and Σ1/n² converges

b_n = \dfrac{1}{n}

The extra factor n^(1/n) approaches 1.

\dfrac{a_n}{b_n} = \dfrac{n}{n \cdot n^{1/n}} = \dfrac{1}{n^{1/n}}

Multiply by n.

\lim_{n \to \infty} \dfrac{1}{n^{1/n}} = 1

Use the hint.

Answer: diverges, since the harmonic series diverges

10.

b_n = \dfrac{n}{n^3} = \dfrac{1}{n^2}

Keep the fastest-growing pieces.

\dfrac{a_n}{b_n} = \dfrac{4n \cdot n^2}{n^3 + 2n + 1} = \dfrac{4n^3}{n^3 + 2n + 1}

Multiply by n².

\lim_{n \to \infty} \dfrac{4n^3}{n^3 + 2n + 1} = 4

Divide the top and bottom by n³.

Answer: yes, the total is finite, since the limit is 4 and Σ1/n² converges

Limit comparison test FAQ

What is the limit comparison test?

It compares two series with positive terms by the limit of the ratio of their terms. If $a_n/b_n$ approaches a positive finite number, both series converge or both diverge.

How do I choose b_n for the limit comparison test?

Keep only the fastest-growing piece of the top and of the bottom, and drop constant factors. For $\frac{3n + 5}{2n^3 - n + 4}$ , that gives $\frac{n}{n^3} = \frac{1}{n^2}$ .

What happens if the limit is 0 or infinity?

If the limit is 0 and $\sum b_n$ converges, then $\sum a_n$ converges. If the limit is ∞ and $\sum b_n$ diverges, then $\sum a_n$ diverges. Other cases give no answer.

When should I use limit comparison instead of direct comparison?

Use it when the terms look like a p-series or geometric series but the inequality goes the wrong way, as with $\frac{1}{n^2 - 1}$ compared with $\frac{1}{n^2}$ .

Summary

If $a_n, b_n \gt 0$ and $a_n/b_n \to c$ with 0 < c < ∞, the two series both converge or both diverge.
Choose $b_n$ from the fastest-growing pieces of the term.
For fractions of powers, $b_n = 1/n^p$ with p = (bottom power) − (top power).
A limit of 0 or ∞ gives an answer only in the two edge cases.
No inequality is needed, which makes it easier than direct comparison for many series.

Previous lesson: Direct Comparison Test

Next lesson: Alternating Series Test

Back to the Calculus 2 Index

Why the ratio of the terms matters

The limit comparison test

How to use the limit comparison test

Limit comparison test examples

Example 1: A term slightly bigger than 1/n²

Example 2: A rational term

Example 3: A square root that acts like n

Example 4: A trig term

Example 5: Forest loss (word problem)

Common mistakes

Related methods

Direct or limit comparison?

Geometric benchmarks

Practice problems

Limit comparison test FAQ

What is the limit comparison test?

How do I choose bn for the limit comparison test?

What happens if the limit is 0 or infinity?

When should I use limit comparison instead of direct comparison?

Summary

How do I choose b_n for the limit comparison test?