Ratio Test

By

The ratio test decides whether a series converges by looking at the ratio of consecutive terms. Find L = \lim_{n \to \infty} \left|\dfrac{a_{n+1}}{a_n}\right|. If L < 1, the series converges absolutely. If L > 1 (or L = ∞), it diverges. If L = 1, the test gives no answer.

Before this lesson: Absolute and Conditional Convergence, Infinite Geometric Series, Permutations and Combinations (for factorials) and Limits at Infinity

Why the ratio of terms matters

In a geometric series, each term is r times the one before, and the series converges exactly when |r| < 1. Many other series act like this in the long run: far out, each term is about L times the one before.

If L is less than 1, the terms eventually shrink at least as fast as a convergent geometric series, so the series converges. If L is more than 1, the terms eventually grow, so they cannot approach 0 and the series diverges.

Dot plot of the ratios of consecutive terms for three series, for n = 1 to 12. For n/3ⁿ (blue circles) the ratios fall toward a dashed line at 1/3. For 1/n² (gray triangles) the ratios climb toward the line at 1. For n!/10ⁿ (orange squares) the ratios rise in a straight line and reach 1 at n = 9 and pass it at n = 10.
The ratio test reads the long-run ratio L: below 1 means convergence, above 1 means divergence, and a ratio that approaches 1 gives no answer.

The ratio test

L = \lim_{n \to \infty} \left|\dfrac{a_{n+1}}{a_n}\right|
  • If L < 1, then \sum a_n converges absolutely (and so it converges).
  • If L > 1 or L = ∞, then \sum a_n diverges.
  • If L = 1, the test is inconclusive. Use a different test.

The ratio test works best when the terms contain factorials like n! or exponentials like 3^n, because those simplify nicely in a ratio. For terms that are just powers of n, such as \frac{1}{n^2}, L is always 1, so use a p-series or comparison test instead.

Factorial facts. (n + 1)! = (n + 1) \cdot n!, so \dfrac{(n + 1)!}{n!} = n + 1. Likewise \dfrac{3^{n+1}}{3^n} = 3.

How to use the ratio test

  1. Write a_{n+1} by replacing every n in a_n with n + 1.
  2. Form \left|\dfrac{a_{n+1}}{a_n}\right| and multiply by the reciprocal of a_n.
  3. Cancel factorials and powers, then find the limit L.
  4. Compare L with 1.

Ratio test examples

Example 1: A power of n over an exponential

Does \sum_{n=1}^{\infty} \dfrac{n}{3^n} converge?

a_{n+1} = \dfrac{n + 1}{3^{n+1}}

Replace n with n + 1.

\dfrac{a_{n+1}}{a_n} = \dfrac{n + 1}{3^{n+1}} \cdot \dfrac{3^n}{n}

Multiply by the reciprocal of aₙ.

\dfrac{n + 1}{3^{n+1}} \cdot \dfrac{3^n}{n} = \dfrac{n + 1}{3n}

Cancel: 3ⁿ⁺¹ ÷ 3ⁿ = 3 (see exponent rules).

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

Divide the top and bottom by n (see limits at infinity).

\dfrac{1}{3} \lt 1

Compare L with 1.

Answer: the series converges absolutely. (Its sum is 3/4.)

Example 2: A factorial in the denominator

Does \sum_{n=0}^{\infty} \dfrac{5^n}{n!} converge?

\dfrac{a_{n+1}}{a_n} = \dfrac{5^{n+1}}{(n + 1)!} \cdot \dfrac{n!}{5^n}

Multiply by the reciprocal of aₙ.

\dfrac{5^{n+1}}{5^n} \cdot \dfrac{n!}{(n + 1)!} = 5 \cdot \dfrac{1}{n + 1}

Group the powers and the factorials; (n + 1)! = (n + 1) · n!.

L = \lim_{n \to \infty} \dfrac{5}{n + 1} = 0

The denominator grows without bound.

0 \lt 1

Compare L with 1.

Answer: the series converges absolutely. Factorials grow faster than any exponential. (Its sum is e⁵ ≈ 148.4.)

Example 3: A factorial in the numerator

Does \sum_{n=1}^{\infty} \dfrac{n!}{10^n} converge?

\dfrac{a_{n+1}}{a_n} = \dfrac{(n + 1)!}{10^{n+1}} \cdot \dfrac{10^n}{n!}

Multiply by the reciprocal of aₙ.

\dfrac{(n + 1)!}{n!} \cdot \dfrac{10^n}{10^{n+1}} = \dfrac{n + 1}{10}

Cancel the factorials and the powers of 10.

L = \lim_{n \to \infty} \dfrac{n + 1}{10} = \infty

The ratio grows without bound.

Answer: the series diverges. The terms shrink at first (0.1, 0.02, 0.006, …), but from n = 10 on each term is at least as big as the one before, so the terms never approach 0 and the series fails the divergence test.

Example 4: When the ratio test fails

Apply the ratio test to \sum_{n=1}^{\infty} \dfrac{1}{n} and to \sum_{n=1}^{\infty} \dfrac{1}{n^2}.

\dfrac{a_{n+1}}{a_n} = \dfrac{1}{n + 1} \cdot \dfrac{n}{1} = \dfrac{n}{n + 1}

For the harmonic series, aₙ = 1/n. Multiply by the reciprocal of aₙ.

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

Divide the top and bottom by n.

\dfrac{a_{n+1}}{a_n} = \dfrac{1}{(n + 1)^2} \cdot \dfrac{n^2}{1} = \left(\dfrac{n}{n + 1}\right)^2

For aₙ = 1/n², multiply by the reciprocal of aₙ.

L = \left(\lim_{n \to \infty} \dfrac{n}{n + 1}\right)^2 = 1^2 = 1

The inside limit is 1, as before.

\text{Both give } L = 1

The test is inconclusive for both.

Answer: the ratio test says nothing here. Yet the first series diverges and the second converges, as the p-series test shows. This is why L = 1 never decides anything.

Example 5: Copies of a file (word problem)

A file-sharing model predicts 100 \cdot \dfrac{n^2}{2^n} new copies of a file on day n. Is the total number of copies over all time finite? (The model’s total turns out to be 600 copies.)

a_n = \dfrac{100n^2}{2^n}

Write the term.

\dfrac{a_{n+1}}{a_n} = \dfrac{100(n + 1)^2}{2^{n+1}} \cdot \dfrac{2^n}{100n^2}

Multiply by the reciprocal of aₙ.

\dfrac{100(n + 1)^2}{2^{n+1}} \cdot \dfrac{2^n}{100n^2} = \dfrac{1}{2}\left(\dfrac{n + 1}{n}\right)^2

Cancel the 100s, and 2ⁿ ÷ 2ⁿ⁺¹ = 1/2.

L = \lim_{n \to \infty} \dfrac{1}{2}\left(1 + \dfrac{1}{n}\right)^2 = \dfrac{1}{2}

1/n goes to 0.

\dfrac{1}{2} \lt 1

Compare L with 1.

Answer: yes, the total converges. New copies peak around day 3 (about 113) and then fall off, so the all-time total levels off at 600 copies.

Common mistakes

MistakeExample of the errorFix
Deciding anything when L = 1“L = 1 for \sum 1/n^2, so it diverges”L = 1 gives no answer. Use another test; here the p-series test shows convergence.
Mishandling factorialsWriting \frac{(n+1)!}{n!} = 1 or (n + 1)! = n! + 1(n + 1)! = (n + 1) \cdot n!, so the ratio is n + 1.
Flipping the ratioUsing \frac{a_n}{a_{n+1}}The test uses \frac{a_{n+1}}{a_n}: the next term over the current one. Flipping it changes L to 1/L.
Forgetting the absolute valueGetting L = −1/2 for \sum (-1)^n/2^n and saying it is less than 1 “by a lot”Take the absolute value: L = 1/2. The sign never affects L.
Replacing only some of the n’sWriting a_{n+1} = \frac{n + 1}{3^n} for a_n = \frac{n}{3^n}Replace every n: a_{n+1} = \frac{n + 1}{3^{n+1}}.
Using it on simple powers of nUsing the ratio test on \frac{3n + 1}{n^3}For terms that are only powers of n, L = 1. Use a comparison test.

Related methods

Ratio test or root test?

The root test, the next lesson, uses \sqrt[n]{|a_n|} instead of a ratio. It is easier when the whole term is raised to the nth power, like \left(\frac{2n}{3n + 1}\right)^n. The ratio test is easier with factorials.

Radius of convergence

The ratio test is the main tool for power series in Unit 7. There, L depends on x, and the condition L < 1 gives the interval where the power series converges.

Practice problems

  1. Does \sum_{n=0}^{\infty} \dfrac{2^n}{n!} converge?
  2. Does \sum_{n=1}^{\infty} \dfrac{n^2}{4^n} converge?
  3. Does \sum_{n=1}^{\infty} \dfrac{3^n}{n^2} converge?
  4. Does \sum_{n=1}^{\infty} \dfrac{n!}{n^n} converge?
  5. Does \sum_{n=1}^{\infty} \dfrac{(-1)^n 4^n}{n \cdot 5^n} converge? If so, how?
  6. What does the ratio test say about \sum_{n=1}^{\infty} \dfrac{2n + 1}{n^3}? Does the series converge?
  7. Does \sum_{n=0}^{\infty} \dfrac{(2n)!}{(n!)^2 \cdot 5^n} converge?
  8. Does \sum_{n=0}^{\infty} \dfrac{10^n}{(2n)!} converge?
  9. Does \sum_{n=1}^{\infty} \dfrac{n^5}{1.01^n} converge?
  10. A video gets 1000 \cdot \dfrac{3^n}{n!} new views on day n, starting at n = 1. Is the total number of views finite? (Hint: \sum_{n=0}^{\infty} 3^n/n! = e^3.)
Show answers

1.

\dfrac{2^{n+1}}{(n + 1)!} \cdot \dfrac{n!}{2^n} = \dfrac{2}{n + 1}

Form the ratio and cancel.

L = \lim_{n \to \infty} \dfrac{2}{n + 1} = 0

Take the limit.

Answer: L = 0 < 1, so it converges absolutely

2.

\dfrac{(n + 1)^2}{4^{n+1}} \cdot \dfrac{4^n}{n^2} = \dfrac{1}{4}\left(\dfrac{n + 1}{n}\right)^2

Form the ratio and cancel.

L = \dfrac{1}{4}

(n + 1)/n goes to 1.

Answer: L = 1/4 < 1, so it converges absolutely

3.

\dfrac{3^{n+1}}{(n + 1)^2} \cdot \dfrac{n^2}{3^n} = 3\left(\dfrac{n}{n + 1}\right)^2

Form the ratio and cancel.

L = 3

n/(n + 1) goes to 1.

Answer: L = 3 > 1, so it diverges

4.

\dfrac{(n + 1)!}{(n + 1)^{n+1}} \cdot \dfrac{n^n}{n!} = \dfrac{(n + 1) n^n}{(n + 1)^{n+1}}

Form the ratio; (n + 1)!/n! = n + 1.

\dfrac{(n + 1) n^n}{(n + 1)^{n+1}} = \dfrac{n^n}{(n + 1)^n} = \left(\dfrac{n}{n + 1}\right)^n

Cancel one factor of n + 1.

\left(\dfrac{n}{n + 1}\right)^n = \dfrac{1}{\left(1 + \frac{1}{n}\right)^n}

Flip the fraction inside.

L = \dfrac{1}{e} \approx 0.368

(1 + 1/n)ⁿ goes to e (see the number e).

Answer: L = 1/e < 1, so it converges absolutely

5.

\left|\dfrac{a_{n+1}}{a_n}\right| = \dfrac{4^{n+1}}{(n + 1) 5^{n+1}} \cdot \dfrac{n \cdot 5^n}{4^n}

Take absolute values; the signs drop out.

\dfrac{4^{n+1}}{(n + 1) 5^{n+1}} \cdot \dfrac{n \cdot 5^n}{4^n} = \dfrac{4}{5} \cdot \dfrac{n}{n + 1}

Cancel.

L = \dfrac{4}{5}

n/(n + 1) goes to 1.

Answer: L = 4/5 < 1, so it converges absolutely

6.

\dfrac{2(n + 1) + 1}{(n + 1)^3} \cdot \dfrac{n^3}{2n + 1} = \dfrac{2n + 3}{2n + 1} \cdot \left(\dfrac{n}{n + 1}\right)^3

Form the ratio.

L = 1 \cdot 1 = 1

Both factors go to 1, so the ratio test is inconclusive.

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

Use the limit comparison test with 1/n² instead.

Answer: the ratio test gives no answer; the series converges by limit comparison with Σ1/n²

7.

\dfrac{(2n + 2)!}{((n + 1)!)^2 5^{n+1}} \cdot \dfrac{(n!)^2 5^n}{(2n)!} = \dfrac{(2n + 2)(2n + 1)}{5(n + 1)^2}

(2n + 2)! = (2n + 2)(2n + 1)(2n)! and ((n + 1)!)² = (n + 1)²(n!)².

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

Cancel one factor: 2n + 2 = 2(n + 1).

L = \lim_{n \to \infty} \dfrac{4n + 2}{5n + 5} = \dfrac{4}{5}

Divide the top and bottom by n.

Answer: L = 4/5 < 1, so it converges absolutely

8.

\dfrac{10^{n+1}}{(2n + 2)!} \cdot \dfrac{(2n)!}{10^n} = \dfrac{10}{(2n + 2)(2n + 1)}

(2n + 2)! = (2n + 2)(2n + 1)(2n)!.

L = 0

The denominator grows without bound.

Answer: L = 0 < 1, so it converges absolutely

9.

\dfrac{(n + 1)^5}{1.01^{n+1}} \cdot \dfrac{1.01^n}{n^5} = \dfrac{1}{1.01}\left(\dfrac{n + 1}{n}\right)^5

Form the ratio and cancel.

L = \dfrac{1}{1.01} \approx 0.990

(n + 1)/n goes to 1.

Answer: L < 1, so it converges (even though the terms first grow for hundreds of days)

10.

\dfrac{3^{n+1}}{(n + 1)!} \cdot \dfrac{n!}{3^n} = \dfrac{3}{n + 1}

Form the ratio and cancel.

L = \lim_{n \to \infty} \dfrac{3}{n + 1} = 0

Take the limit. L is less than 1, so the total converges.

1000\sum_{n=1}^{\infty} \dfrac{3^n}{n!} = 1000\left(e^3 - 1\right)

The n = 0 term of the hint is 1, so subtract it.

1000\left(e^3 - 1\right) \approx 19{,}086

e³ ≈ 20.086.

Answer: yes, the total is finite: about 19,086 views

Ratio test FAQ

What does the ratio test say?

Find L, the limit of |aₙ₊₁/aₙ|. If L < 1, the series converges absolutely; if L > 1, it diverges; if L = 1, the test gives no answer.

When should I use the ratio test?

Use it when the terms contain factorials or exponentials like 2^n or n^n. These cancel neatly in the ratio. Avoid it for terms built only from powers of n.

Why is the ratio test inconclusive when L = 1?

Series with L = 1 can go either way. Both \sum 1/n and \sum 1/n^2 give L = 1, but the first diverges and the second converges.

Does the ratio test show absolute or conditional convergence?

When L < 1, it shows absolute convergence, because it uses the absolute values of the terms. It can never show conditional convergence.

Summary

  • Find L = lim |aₙ₊₁/aₙ|.
  • L < 1: converges absolutely. L > 1 or ∞: diverges. L = 1: no answer.
  • Use (n + 1)! = (n + 1) \cdot n! and a^{n+1}/a^n = a to simplify.
  • Best for factorials and exponentials; useless for plain powers of n.
  • The ratio test is the key tool for power series later on.

Previous lesson: Absolute and Conditional Convergence

Next lesson: Root Test

Back to the Calculus 2 Index