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Ratio Test

Ratio Test

A convergence test used when terms of a series contain factorials and/or nth powers.

 

Ratio Test: R = lim|a_(n+1)/a_n|. If R<1, series converges absolutely; R>1, diverges; R=1, inconclusive.

 

 

See also

Root test, convergent series, divergent series, absolute convergence, limit, infinity

Key Formula

R=lim⁡n→∞∣an+1an∣R = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|
Where:
  • RR = The limit of the absolute ratio of consecutive terms
  • ana_n = The nth term of the series
  • an+1a_{n+1} = The (n+1)th term of the series

Worked Example

Problem: Determine whether the series ∑n=1∞3nn!\displaystyle\sum_{n=1}^{\infty} \frac{3^n}{n!} converges or diverges.
Step 1: Identify the general term of the series.
an=3nn!a_n = \frac{3^n}{n!}
Step 2: Write the (n+1)th term by replacing every n with n+1.
an+1=3n+1(n+1)!a_{n+1} = \frac{3^{n+1}}{(n+1)!}
Step 3: Form the ratio and simplify. Recall that (n+1)!=(n+1)⋅n!(n+1)! = (n+1) \cdot n!.
∣an+1an∣=3n+1(n+1)!⋅n!3n=3n+1\left| \frac{a_{n+1}}{a_n} \right| = \frac{3^{n+1}}{(n+1)!} \cdot \frac{n!}{3^n} = \frac{3}{n+1}
Step 4: Take the limit as n approaches infinity.
R=lim⁡n→∞3n+1=0R = \lim_{n \to \infty} \frac{3}{n+1} = 0
Step 5: Apply the Ratio Test conclusion. Since R=0<1R = 0 < 1, the series converges.
Answer: The series ∑3nn!\sum \frac{3^n}{n!} converges by the Ratio Test because R=0<1R = 0 < 1.

Another Example

This example shows a divergent case where the factorial grows faster than the exponential, producing a limit greater than 1. It contrasts with the first example where the exponential was in the numerator and the factorial in the denominator.

Problem: Determine whether the series ∑n=1∞n!2n\displaystyle\sum_{n=1}^{\infty} \frac{n!}{2^n} converges or diverges.
Step 1: Identify the general term.
an=n!2na_n = \frac{n!}{2^n}
Step 2: Write the next term.
an+1=(n+1)!2n+1a_{n+1} = \frac{(n+1)!}{2^{n+1}}
Step 3: Form the ratio and simplify. Note that (n+1)!=(n+1)⋅n!(n+1)! = (n+1) \cdot n! and 2n+1=2⋅2n2^{n+1} = 2 \cdot 2^n.
∣an+1an∣=(n+1)!2n+1⋅2nn!=n+12\left| \frac{a_{n+1}}{a_n} \right| = \frac{(n+1)!}{2^{n+1}} \cdot \frac{2^n}{n!} = \frac{n+1}{2}
Step 4: Take the limit as n approaches infinity.
R=lim⁡n→∞n+12=∞R = \lim_{n \to \infty} \frac{n+1}{2} = \infty
Step 5: Since R=∞>1R = \infty > 1, the series diverges by the Ratio Test.
Answer: The series ∑n!2n\sum \frac{n!}{2^n} diverges by the Ratio Test because R=∞>1R = \infty > 1.

Frequently Asked Questions

What happens when the Ratio Test gives R = 1?
When R=1R = 1, the Ratio Test is inconclusive — it cannot determine whether the series converges or diverges. You must use a different test, such as the Root Test, Comparison Test, or Integral Test, to settle the question. Classic examples where R=1R = 1 include the harmonic series ∑1n\sum \frac{1}{n} (diverges) and the p-series ∑1n2\sum \frac{1}{n^2} (converges), showing that R=1R = 1 can go either way.
When should you use the Ratio Test instead of the Root Test?
Use the Ratio Test when your series involves factorials (like n!n!) because the ratio of consecutive factorials simplifies cleanly: (n+1)!/n!=n+1(n+1)!/n! = n+1. The Root Test is often better when terms involve only nth powers, such as (23)n\left(\frac{2}{3}\right)^n, because taking the nth root removes the exponent directly. Both tests give the same result when both apply, so it comes down to which simplifies more easily.
Does the Ratio Test prove absolute convergence?
Yes. The Ratio Test uses absolute values, so when it concludes convergence (R<1R < 1), it actually proves absolute convergence. This is a stronger result than ordinary convergence because any absolutely convergent series is also convergent.

Ratio Test vs. Root Test

Ratio TestRoot Test
FormulaR=lim⁡n→∞∣an+1an∣R = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|L=lim⁡n→∞∣an∣nL = \lim_{n \to \infty} \sqrt[n]{|a_n|}
Best used whenTerms contain factorials or products of consecutive integersTerms are raised to the nth power, like (f(n))n(f(n))^n
Convergence criterionR<1R < 1: converges; R>1R > 1: diverges; R=1R = 1: inconclusiveL<1L < 1: converges; L>1L > 1: diverges; L=1L = 1: inconclusive
Proves absolute convergence?YesYes
Key advantageFactorials cancel neatly in the rationth powers simplify neatly under the nth root

Why It Matters

The Ratio Test appears throughout calculus II and is one of the most frequently tested convergence tests on AP Calculus BC and college-level exams. It is the standard tool for finding the radius of convergence of a power series, where you apply the test with the variable xx included and solve the resulting inequality. Mastering it also builds intuition for how different growth rates (factorial vs. exponential vs. polynomial) determine whether a series converges.

Common Mistakes

Mistake: Forgetting the absolute value in the ratio and getting a negative limit.
Correction: Always use ∣an+1an∣\left| \frac{a_{n+1}}{a_n} \right|. The Ratio Test compares magnitude, so you must take the absolute value of the ratio before evaluating the limit.
Mistake: Trying to draw a conclusion when R = 1.
Correction: When R=1R = 1, the test is inconclusive. Do not claim convergence or divergence — switch to another test. For example, the harmonic series gives R=1R = 1 but diverges, while ∑1/n2\sum 1/n^2 also gives R=1R = 1 but converges.

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