Ratio Test for Convergence

AP Calculus BC· AP Calculus BC CED — Infinite Sequences and Series· 14 min read

1. Core Concepts of the Ratio Test★★☆☆☆⏱ 4 min

The Ratio Test (also called d'Alembert's Ratio Test) is a convergence test for infinite series $\sum_{n = 1}^{\infty} a_{n}$ , most useful when series terms contain factorials, exponential functions, or powers of $n$ that simplify cleanly when taking the ratio of consecutive terms. This topic is tested on both multiple-choice and free-response sections of the AP Calculus BC exam.

📘 Definition

Ratio Test

L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|

A convergence test that compares the asymptotic growth rate of consecutive terms of an infinite series. Three outcomes: 1) If $L < 1$ , the series converges absolutely; 2) If $L > 1$ (including $L = \infty$ ), the series diverges; 3) If $L = 1$ , the test is inconclusive.

Example:

The limit $L$ behaves like the absolute value of the common ratio for geometric series.

📐 Worked Example

Determine whether the series $\sum_{n = 1}^{\infty} \frac{3 ^{n}}{n !}$ converges or diverges.

1
Identify $a_{n}$ and $a_{n + 1}$ :
$a_n = \frac{3^n}{n!}, \quad a_{n+1} = \frac{3^{n+1}}{(n+1)!}$
2
Set up the ratio of consecutive terms (all terms are positive, so absolute value can be dropped):
$\frac{a_{n+1}}{a_n} = \frac{3^{n+1}}{(n+1)!} \cdot \frac{n!}{3^n}$
3
Simplify the expression:
$\frac{3^{n+1}}{3^n} = 3, \quad \frac{n!}{(n+1)!} = \frac{1}{n+1} \implies \frac{a_{n+1}}{a_n} = \frac{3}{n+1}$
4
Compute the limit as $n \to \infty$ :
$L = \lim_{n \to \infty} \frac{3}{n+1} = 0$
5
Apply the ratio test rule: $0 < 1$ , so the series converges absolutely.

2. Inconclusive Cases★★★☆☆⏱ 3 min

When $L = 1$ , the ratio test gives no information about convergence, and this is a common testing point on the AP exam. The ratio test only captures exponential growth rates of terms; when growth is polynomial, the limit $L$ will always equal 1, regardless of whether the series converges or diverges. Whenever you calculate $L = 1$ , you must switch to another appropriate test.

📐 Worked Example

The ratio test for the series $\sum_{n = 1}^{\infty} \frac{n}{2 n ^{2} + 1}$ gives $L = 1$ . Determine whether the series converges or diverges.

1
Confirm that $L = 1$ means the ratio test is inconclusive, so we use the limit comparison test.
2
For large $n$ , the end behavior of $a_{n}$ is approximately $\frac{n}{2 n ^{2}} = \frac{1}{2 n}$ , so we compare to the divergent series $\sum \frac{1}{n}$ .
3
Compute the limit of the ratio of terms:
$\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{n/(2n^2 +1)}{1/n} = \lim_{n \to \infty} \frac{n^2}{2n^2 +1} = \frac{1}{2}$
4
Since the limit is positive and finite, the original series diverges by the limit comparison test.

3. Applying the Ratio Test to Power Series★★★☆☆⏱ 4 min

The most common high-stakes use of the ratio test on the AP Calculus BC exam is finding the radius and interval of convergence for a power series of the form $\sum_{n = 0}^{\infty} c_{n} (x - a)^{n}$ . Convergence of a power series depends on the value of $x$ , and the ratio test is ideal here because the $(x - a)$ term simplifies cleanly when taking the ratio of consecutive terms.

Write $a_{n} = c_{n} (x - a)^{n}$ , then set up the ratio $\frac{a _{n + 1}}{a _{n}}$
Separate terms involving $x$ from terms involving $n$
Compute the limit $L (x) = ∣ x - a ∣ \cdot lim_{n \to \infty} \frac{c _{n + 1}}{c _{n}}$
Set $L (x) < 1$ to find the radius of convergence $R = 1/ (lim_{n \to \infty} \frac{c _{n + 1}}{c _{n}})$
Check the endpoints $x = a \pm R$ separately with another test

📐 Worked Example

Find the radius of convergence for the power series $\sum_{n = 1}^{\infty} \frac{( 3 x ) ^{n}}{n 4 ^{n}}$ .

1
Identify $a_{n}$ and $a_{n + 1}$ :
$a_n = \frac{(3x)^n}{n 4^n}, \quad a_{n+1} = \frac{(3x)^{n+1}}{(n+1) 4^{n+1}}$
2
Set up and simplify the absolute value ratio:
$\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(3x)^{n+1}}{(n+1) 4^{n+1}} \cdot \frac{n 4^n}{(3x)^n} \right| = \frac{3|x|}{4} \cdot \frac{n}{n+1}$
3
Compute the limit as $n \to \infty$ :
$\lim_{n \to \infty} \frac{n}{n+1} = 1 \implies L = \frac{3|x|}{4}$
4
Set $L < 1$ for convergence:
$\frac{3|x|}{4} < 1 \implies |x| < \frac{4}{3}$
5
The radius of convergence is $R = \frac{4}{3}$ .

4. AP-Style Concept Check★★★★☆⏱ 3 min

✓ Quick check

Test your understanding with this multiple-choice question:

Which of the following is the radius of convergence of the power series $\sum_{n = 0}^{\infty} \frac{n ! x ^{n}}{4 ^{n}}$ ?
- 0
- 4
- $\frac{1}{4}$
- $\infty$

5. Common Pitfalls

Wrong move:

Reversing the ratio order, computing $\frac{a _{n}}{a _{n + 1}}$ instead of $\frac{a _{n + 1}}{a _{n}}$ , leading to an inverted $L$ and wrong conclusion.

Why:

Students mix up the order when writing quickly, especially for power series problems.

Correct move:

Always explicitly label $a_{n}$ and $a_{n + 1}$ first, then write $a_{n + 1}$ in the numerator and $a_{n}$ in the denominator before simplifying.

Wrong move:

Forgetting to take the absolute value of the ratio, leading to a negative $L$ and incorrect conclusion that $L < 1$ even for divergent series.

Why:

Students drop the absolute value for positive-term series and forget the habit for alternating/power series.

Correct move:

Always write the absolute value around the ratio before simplifying, even if all terms are positive, to build consistency.

Wrong move:

Concluding convergence or divergence when $L = 1$ , without using a second test.

Why:

Students assume the ratio test can answer all questions and forget the inconclusive rule.

Correct move:

As soon as you get $L = 1$ , write "The ratio test is inconclusive" on your paper, then select another appropriate test.

Wrong move:

Leaving $x$ inside the limit when calculating radius of convergence, leading to incorrect simplification.

Why:

Students forget $x$ is a constant with respect to the limit over $n$ .

Correct move:

Always factor all terms involving $x$ out of the limit before evaluating.

Wrong move:

Treating $L = \infty$ as an inconclusive case instead of a divergent case.

Why:

Students only remember $L < 1$ and $L = 1$ and forget that infinite $L$ falls into the $L > 1$ category.

Correct move:

If the limit goes to infinity, note that $\infty > 1$ , so the series diverges by the ratio test.

6. Quick Reference Cheatsheet

Category	Formula/Rule	Notes
Core Ratio Test Limit	$L = lim_{n \to \infty} \frac{a _{n + 1}}{a _{n}}$	For any infinite series $\sum a_{n}$ ; $a_{n + 1}$ always in numerator
Convergence Case	$L < 1$	Series converges absolutely, so it converges
Divergence Case	$L > 1$ (includes $L = \infty$ )	Series diverges, $lim_{n \to \infty} a_{n} \neq = 0$
Inconclusive Case	$L = 1$	Use another test: p-test, limit comparison, alternating series test
Radius of Convergence	$R = \frac{1}{l i m _{n \to \infty} ∣ \frac{c _{n + 1}}{c _{n}} ∣}$	Check endpoints $x = a \pm R$ separately
Infinite Radius	$R = \infty$ when $lim_{n \to \infty} \frac{c _{n + 1}}{c _{n}} = 0$	Converges for all real $x$
Zero Radius	$R = 0$ when $lim_{n \to \infty} \frac{c _{n + 1}}{c _{n}} = \infty$	Converges only at center $x = a$
Power Series Rule	Converges for $∣ x - a ∣ < R$ , diverges for $∣ x - a ∣ > R$	Interval of convergence only includes endpoints if they pass a separate test

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · BC
Find radius of convergence
2022 · BC
Test series convergence

What's Next

Mastering the ratio test is a critical prerequisite for finding the full interval of convergence of power series, which is a 3-5 point question on almost every AP Calculus BC FRQ section. Without correctly applying the ratio test to find the radius of convergence, you cannot complete the interval of convergence, so this topic is make-or-break for a high Unit 10 score. The ratio test also prepares you for the root test, another convergence test, and underpins analysis of Taylor and Maclaurin series, the core of the second half of Unit 10.