Ratio test for convergence — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: The general ratio test formula for infinite series, identifying absolute convergence, conditional convergence, and divergence, handling inconclusive cases, and applying the test to find the radius of convergence of power series.

You should already know: How to compute limits of ratios of sequences, definition of absolute and conditional convergence, basics of infinite series convergence.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus BC style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Ratio test for convergence?

The Ratio Test (also called d'Alembert's Ratio Test, the common synonym in calculus curricula) 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. According to the AP Calculus BC Course and Exam Description (CED), this topic is part of Unit 10: Infinite Sequences and Series, which accounts for 17-18% of the total AP exam score. The ratio test appears in both multiple-choice (MCQ) and free-response (FRQ) sections: it is tested as a standalone convergence check in MCQ, and as a core tool to find the interval of convergence for power series, a high-frequency FRQ question. The core idea of the ratio test is to compare the asymptotic growth rate of consecutive terms: if each subsequent term grows more slowly than a convergent geometric series, the full series converges; if it grows faster than a divergent geometric series, the full series diverges.

2. The Core Ratio Test Rule

The ratio test relies on computing the limit $L$ of the absolute value of the ratio of the $(n+1)$th term to the $n$th term as $n$ approaches infinity. The formal rule is: For any infinite series ${\sum }_{n=1}^{\infty }{a}_n$, compute $$L=\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|$$ Three cases follow from the value of $L$:

1. If $L<1$, the series $\sum a_n$ converges absolutely (and therefore converges, since absolute convergence implies convergence).
2. If $L>1$ (including the case $L=\infty$), the series $\sum a_n$ diverges, because ${\lim }_{n\to \infty }{a}_n\ne 0$, a requirement for any convergent series.
3. If $L=1$, the ratio test is inconclusive: the series could converge or diverge, and you must use another test to reach a conclusion.

The intuition matches geometric series, the simplest case of an infinite series: the limit $L$ acts like the absolute value of the common ratio of a geometric series. Just as a geometric series converges when $|r|<1$ and diverges when $|r|>1$, the same logic extends to any series whose terms behave like a geometric series for large $n$.

Worked Example

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

1. Identify $a_n$ and $a_{n+1}$: $a_n=\frac{3^n}{n!}$, so $a_{n+1}=\frac{3^{n+1}}{(n+1)!}$.
2. Set up the ratio of consecutive terms: since all terms are positive, the absolute value can be dropped, giving $\frac{a_{n+1}}{a_n}=\frac{3^{n+1}}{(n+1)!}\cdot \frac{n!}{3^n}$.
3. Simplify the expression: $\frac{3^{n+1}}{3^n}=3$, and $\frac{n!}{(n+1)!}=\frac{1}{n+1}$, so the ratio simplifies to $\frac{3}{n+1}$.
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.

Exam tip: Always take the absolute value of the ratio before computing the limit. This ensures you correctly test for absolute convergence, even for alternating series and power series with negative $x$ values.

3. Inconclusive Cases of the Ratio Test

When $L=1$, the ratio test gives no information about convergence, and this is a common point of testing on the AP exam, where questions often ask you to identify when the test is inconclusive or what step to take next. Why does this happen? 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. For example, both the convergent p-series $\sum \frac{1}{n^2}$ ($p=2>1$) and the divergent harmonic series $\sum \frac{1}{n}$ ($p=1$) give $L=1$ when the ratio test is applied, so the test cannot distinguish between them. Whenever you calculate $L=1$, you must switch to another test appropriate for the series: common choices are the p-series test for polynomial terms, the limit comparison test for positive-term series, or the alternating series test for alternating series.

Worked Example

Problem: The ratio test for the series ${\sum }_{n=1}^{\infty }\frac{n}{2n^2+1}$ 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 leading terms of the numerator and denominator give end behavior $a_n=\frac{n}{2n^2+1}\approx \frac{n}{2n^2}=\frac{1}{2n}$, so we compare to the divergent series $\sum b_n=\sum \frac{1}{n}$.
3. Compute the limit of the ratio of terms: ${\lim }_{n\to \infty }\frac{a_n}{b_n}={\lim }_{n\to \infty }\frac{n/(2n^2+1)}{1/n}={\lim }_{n\to \infty }\frac{n^2}{2n^2+1}=\frac{1}{2}$, which is a positive finite number.
4. By the limit comparison test, since $\sum \frac{1}{n}$ diverges, the original series also diverges.

Exam tip: If you get $L=1$ on the AP exam, never stop and conclude convergence or divergence. You must explicitly state the ratio test is inconclusive and use a second test to earn full credit.

4. Applying the Ratio Test to Find Radius of Convergence

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. For any power series, the process is:

1. Write $a_n=c_n(x-a{)}^n$, then set up the ratio $\left|\frac{a_{n+1}}{a_n}\right|$.
2. Separate terms involving $x$ from terms involving $n$, so $\left|\frac{a_{n+1}}{a_n}\right|=|x-a|\cdot \left|\frac{c_{n+1}}{c_n}\right|$.
3. Compute the limit $L(x)=|x-a|\cdot {\lim }_{n\to \infty }\left|\frac{c_{n+1}}{c_n}\right|$.
4. Set $L(x)<1$ to find the interval of $x$ where the series converges absolutely, giving the radius of convergence $R=1/\left({\lim }_{n\to \infty }\left|\frac{c_{n+1}}{c_n}\right|\right)$.
5. Check the endpoints $x=a\pm R$ separately with another test, since the ratio test will always be inconclusive at endpoints.

Worked Example

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

1. Identify $a_n=\frac{(3x{)}^n}{n{4}^n}$, so $a_{n+1}=\frac{(3x{)}^{n+1}}{(n+1)4^{n+1}}$.
2. Set up and simplify the 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 $n\to \infty$: ${\lim }_{n\to \infty }\frac{n}{n+1}=1$, so $L=\frac{3|x|}{4}$.
4. Set $L<1$ for convergence: $\frac{3|x|}{4}<1⟹|x|<\frac{4}{3}$.
5. The radius of convergence is $R=\frac{4}{3}$.

Exam tip: AP FRQ questions almost always require endpoint checks after finding the radius with the ratio test. Never skip checking endpoints if the question asks for the full interval of convergence.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Reversing the ratio order, computing $\left|\frac{a_n}{a_{n+1}}\right|$ instead of $\left|\frac{a_{n+1}}{a_n}\right|$, 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. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

Which of the following is the radius of convergence of the power series ${\sum }_{n=0}^{\infty }\frac{n!x^n}{4^n}$? A) $0$ B) $4$ C) $\frac{1}{4}$ D) $\infty$

Worked Solution: First, apply the ratio test for power series. We have $a_n=\frac{n!x^n}{4^n}$, so $a_{n+1}=\frac{(n+1)!x^{n+1}}{4^{n+1}}$. The ratio simplifies to $\left|\frac{a_{n+1}}{a_n}\right|=\frac{|x|(n+1)}{4}$. Taking the limit as $n\to \infty$ gives $L=\infty$ for any $x\ne 0$. The only value of $x$ where $L<1$ is $x=0$, so the radius of convergence is $0$. The correct answer is A.

Question 2 (Free Response)

Consider the series ${\sum }_{n=1}^{\infty }\frac{(-1{)}^{n}n(x+2{)}^n}{3^n}$. (a) Use the ratio test to find the radius of convergence of the series. (b) Find all values of $x$ for which the ratio test is inconclusive. (c) Determine the full interval of convergence of the series.

Worked Solution: (a) We have $a_n=\frac{(-1{)}^{n}n(x+2{)}^n}{3^n}$, so the ratio simplifies to: $$\left|\frac{a_{n+1}}{a_n}\right|=\frac{|x+2|}{3}\cdot \frac{n+1}{n}$$ Taking the limit as $n\to \infty$ gives $L=\frac{|x+2|}{3}$. Setting $L<1$ gives $|x+2|<3$, so the radius of convergence $R=3$.

(b) The ratio test is inconclusive when $L=1$, so $\frac{|x+2|}{3}=1⟹|x+2|=3$. This gives the values $x=1$ and $x=-5$.

(c) Test the endpoints:

• At $x=1$: Substitute to get ${\sum }_{n=1}^{\infty }(-1{)}^{n}n$, which diverges by the nth term test (${\lim }_{n\to \infty }|(-1{)}^{n}n|=\infty \ne 0$).
• At $x=-5$: Substitute to get ${\sum }_{n=1}^{\infty }(-1{)}^{n}n(-3{)}^n/3^n=\sum n$, which also diverges by the nth term test. The full interval of convergence is $(-5,1)$.

Question 3 (Application / Real-World Style)

An economist modeling the total long-term output of a new industry estimates that total cumulative output $Y$ (in millions of dollars) after $t$ years is given by the infinite series $Y={\sum }_{n=1}^{\infty }\frac{(0.6t{)}^n}{n!}$, where $t>0$. For what values of $t$ does this series converge to a finite total output?

Worked Solution: Apply the ratio test with $a_n=\frac{(0.6t{)}^n}{n!}$. The ratio simplifies to $\left|\frac{a_{n+1}}{a_n}\right|=\frac{0.6t}{n+1}$. Taking the limit as $n\to \infty$ gives $L={\lim }_{n\to \infty }\frac{0.6t}{n+1}=0$ for any finite $t>0$. Since $L=0<1$, the series converges for all finite positive values of $t$. In context, this means the model predicts a finite total cumulative output for any finite time horizon, which aligns with expected long-term industry growth behavior.

7. Quick Reference Cheatsheet

8. 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 for series with nth powers, and underpins the analysis of Taylor and Maclaurin series, the core of the second half of Unit 10. Many AP exam questions connect ratio test results to error bounds for Taylor polynomials, so a solid foundation here pays off across multiple question types.

• Interval of convergence
• Root test for convergence
• Maclaurin series
• Taylor series