Radius and interval of convergence of power series — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Finding radius of convergence via the Ratio Test and Root Test, testing series convergence at interval endpoints, identifying absolute/conditional convergence, and handling edge cases of zero or infinite radius of convergence.

You should already know: How to apply the Ratio Test for infinite series, how to test for conditional and absolute convergence of series, basic limit evaluation techniques.

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 Radius and interval of convergence of power series?

A power series is an infinite series of the form ${\sum }_{n=0}^{\infty }{c}_n(x-a{)}^n$, where $a$ is the center of the series and $c_n$ are constant coefficients. For any input $x$, a power series reduces to a numerical infinite series that either converges or diverges. The set of all $x$ for which the series converges is always an interval centered at $a$, called the interval of convergence. The radius of convergence $R$ is half the length of this interval; it measures how far from the center $a$ the series will converge. If the series converges only at $x=a$, $R=0$; if it converges for all real $x$, $R=\infty$, and the interval of convergence spans the entire real line. Per the AP Calculus BC CED, content from the infinite sequences and series unit (including this topic) makes up 17-18% of the total exam score. This topic appears in both multiple-choice (MCQ) and free-response (FRQ) sections, and is almost always paired with other series topics like Taylor series or integral approximation.

2. Finding Radius of Convergence

The standard method for finding $R$ on the AP exam is the Ratio Test, which works seamlessly for the factorial, polynomial, and exponential terms common in Taylor and Maclaurin power series. Recall that for any series $\sum a_n$, the Ratio Test calculates $L={\lim }_{n\to \infty }\left|\frac{a_{n+1}}{a_n}\right|$: the series converges absolutely if $L<1$, diverges if $L>1$, and is inconclusive if $L=1$. For a power series ${\sum }_{n=0}^{\infty }{c}_n(x-a{)}^n$, the general term is $a_n=c_n(x-a{)}^n$. Substituting into the Ratio Test gives: $$L=\underset{n\to \infty }{\lim }\left|\frac{c_{n+1}(x-a{)}^{n+1}}{c_n(x-a{)}^n}\right|=|x-a|\cdot \underset{n\to \infty }{\lim }\left|\frac{c_{n+1}}{c_n}\right|$$ For convergence, we require $L<1$, which rearranges to the radius of convergence formula: $$R=\frac{1}{{\lim }_{n\to \infty }\left|\frac{c_{n+1}}{c_n}\right|}$$ For series with terms raised to the nth power, the Root Test is an alternative: $R=1/\left({\lim }_{n\to \infty }\sqrt[n]{|c_n|}\right)$, but this is rarely tested on the AP exam. The intuition for $R$ is straightforward: it is the maximum distance from the center where the series terms decay fast enough for convergence. If the limit of the ratio is 0, $R=\infty$ (converges everywhere); if the limit is infinite, $R=0$ (converges only at the center).

Worked Example

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

1. Identify the general term: $a_n=\frac{3^n(x-4{)}^n}{n!}$, so $a_{n+1}=\frac{3^{n+1}(x-4{)}^{n+1}}{(n+1)!}$.
2. Simplify the ratio of absolute values: $$\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{3^{n+1}(x-4{)}^{n+1}}{(n+1)!}\cdot \frac{n!}{3^n(x-4{)}^n}\right|=\frac{3|x-4|}{n+1}$$
3. Evaluate the limit as $n\to \infty$: $L={\lim }_{n\to \infty }\frac{3|x-4|}{n+1}=0$ for any finite $x$.
4. Since $L=0<1$ for all $x$, $R=1/0=\infty$.

Exam tip: Always factor $|x-a|$ out of the limit before evaluating—this term does not depend on $n$, so factoring it out simplifies your limit calculation and avoids algebraic errors.

3. Finding Interval of Convergence: Testing Endpoints

Once you have the radius of convergence $R$, you know the open interval of convergence is $(a-R,a+R)$: for any $x$ inside this open interval, the Ratio Test guarantees absolute convergence. However, at the two endpoints $x=a-R$ and $x=a+R$, $|x-a|=R$, which means $L=1$, and the Ratio Test is inconclusive. You must test convergence at each endpoint separately using other convergence tests: the nth-Term Test for Divergence, Alternating Series Test, p-Series Test, or Comparison Test, depending on the form of the series at the endpoint. At each endpoint, the series can converge absolutely, converge conditionally, or diverge. You include any endpoint that converges (either absolutely or conditionally) in your final interval of convergence. The AP exam explicitly tests whether you remember to check endpoints—omitting this step will cost you points on FRQ, and most MCQ distractors are wrong intervals that miss correct endpoint inclusion.

Worked Example

Find the full interval of convergence for the power series ${\sum }_{n=1}^{\infty }\frac{(x+2{)}^n}{n\cdot 3^n}$, given that its radius of convergence is $R=3$.

1. Calculate the open interval: center $a=-2$, so the open interval is $(-2-3,-2+3)=(-5,1)$.
2. Test the left endpoint $x=-5$: substitute $x=-5$ into the series: $$\sum _{n=1}^{\infty }\frac{(-5+2{)}^n}{n\cdot 3^n}=\sum _{n=1}^{\infty }\frac{(-3{)}^n}{n\cdot 3^n}=\sum _{n=1}^{\infty }\frac{(-1{)}^n}{n}$$ This is the alternating harmonic series, which converges by the Alternating Series Test, so include $x=-5$.
3. Test the right endpoint $x=1$: substitute $x=1$ into the series: $$\sum _{n=1}^{\infty }\frac{(1+2{)}^n}{n\cdot 3^n}=\sum _{n=1}^{\infty }\frac{3^n}{n\cdot 3^n}=\sum _{n=1}^{\infty }\frac{1}{n}$$ This is the harmonic p-series with $p=1$, which diverges, so exclude $x=1$.
4. Final interval of convergence: $[-5,1)$.

Exam tip: When testing endpoints, simplify the series fully before applying a convergence test—$R^n$ almost always cancels out completely, leaving you with a simple alternating or positive series that is easy to test.

4. Edge Cases for Convergence

Two edge cases appear regularly on the AP exam, and both are common sources of lost points. The first edge case is convergence only at the center, which gives $R=0$ and an interval of convergence that is just the single point $\{a\}$. This occurs when the coefficients $c_n$ grow so quickly that for any $x\ne a$, the limit $L$ from the Ratio Test is greater than 1, so the series diverges everywhere except the center. The second edge case is convergence for all real $x$, which gives $R=\infty$ and an interval of convergence of $(-\infty ,\infty )$. This occurs when the coefficients decay very quickly (most commonly when denominators have factorials, as in the Maclaurin series for $e^x$, $\sin x$, and $\cos x$), so the limit $L$ is 0 for any $x$, which is always less than 1. A third less common edge case is a power series with non-zero coefficients only for even or odd powers of $(x-a)$; the method for finding $R$ does not change, but you will get an extra factor of $(x-a{)}^2$ in your ratio, so be careful to simplify correctly.

Worked Example

Find the radius and interval of convergence for the power series ${\sum }_{n=0}^{\infty }n!(4x-8{)}^n$.

1. Rewrite the series in standard form: ${\sum }_{n=0}^{\infty }n!4^n(x-2{)}^n$, so center $a=2$, $c_n=n!4^n$.
2. Apply the Ratio Test: $$\underset{n\to \infty }{\lim }\left|\frac{(n+1)!4^{n+1}(x-2{)}^{n+1}}{n!4^n(x-2{)}^n}\right|=\underset{n\to \infty }{\lim }4(n+1)|x-2|$$
3. For any $x\ne 2$, this limit is $\infty >1$, so the series diverges for all $x\ne 2$.
4. At $x=2$, the series becomes ${\sum }_{n=0}^{\infty }n!(0{)}^n=1+0+0+...=1$, which converges.
5. Final result: radius of convergence $R=0$, interval of convergence is $\{2\}$ (or $[2,2]$).

Exam tip: Always explicitly confirm convergence at the center for $R=0$ cases—remember that every power series converges at its center, even if it diverges everywhere else.

5. Common Pitfalls (and how to avoid them)

• Wrong move: After finding the radius of convergence $R$, stopping and writing the interval as $(a-R,a+R)$ without testing endpoints. Why: Students assume endpoints always diverge or always converge, but AP exam problems are designed to have one or both endpoints converge, explicitly testing this step. Correct move: Always substitute both endpoints into the original series and test for convergence with an appropriate non-Ratio Test, even if you think you know the result.
• Wrong move: Forgetting the absolute value when applying the Ratio Test, leading to a negative limit and an incorrect negative radius. Why: Students drop the absolute value out of habit when working with positive terms, but $(x-a)$ can be negative. Correct move: Always write the absolute value around the entire ratio before simplifying, and keep it until you have isolated $|x-a|$.
• Wrong move: When the power series is given in the form $\sum c_n(kx-a{)}^n$, failing to factor out $k$ before calculating $R$, leading to an incorrect interval. Why: Students treat $kx-a$ the same as $x-a$ without adjusting for the coefficient of $x$. Correct move: Rewrite the series to standard form $\sum c_n{k}^n(x-a/k{)}^n$ before calculating the limit for $R$, so you correctly identify the center and coefficients.
• Wrong move: When $L=1$ at an endpoint, concluding that the endpoint diverges because the Ratio Test is inconclusive. Why: Students confuse "inconclusive" with "divergent"—the Ratio Test just does not give an answer, not that the answer is divergence. Correct move: If the Ratio Test is inconclusive at an endpoint, always use another test (p-Series, Alternating Series Test, Comparison Test) to determine convergence.
• Wrong move: Claiming the interval of convergence is $(-R,R)$ when the center is not 0, forgetting to shift the interval to the center. Why: Students are used to working with Maclaurin series (center 0) and forget to adjust for Taylor series centered at a non-zero point. Correct move: Always write the interval as $(a-R,a+R)$ where $a$ is the given center before testing endpoints.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

What is the radius of convergence of the power series ${\sum }_{n=0}^{\infty }\frac{(2x{)}^n}{(n+1{)}^3}$? A) $\frac{1}{2}$ B) $1$ C) $2$ D) $\infty$

Worked Solution: Rewrite the series in standard form as ${\sum }_{n=0}^{\infty }\frac{2^n}{(n+1{)}^3}{x}^n$, so $c_n=2^n/(n+1{)}^3$. Apply the Ratio Test: ${\lim }_{n\to \infty }\left|\frac{2^{n+1}/(n+2{)}^3{x}^{n+1}}{2^n/(n+1{)}^3{x}^n}\right|=2|x|{\lim }_{n\to \infty }\frac{(n+1{)}^3}{(n+2{)}^3}=2|x|(1)=2|x|$. For convergence, we require $2|x|<1$, so $|x|<1/2$, which means $R=1/2$. The correct answer is A.

Question 2 (Free Response)

Consider the power series ${\sum }_{n=1}^{\infty }\frac{(-1{)}^n(x-1{)}^n}{n\cdot 2^n}$. (a) Find the radius of convergence of the series. (b) Find the interval of convergence of the series. (c) For what values of $x$ does the series converge absolutely?

Worked Solution: (a) Let $a_n=\frac{(-1{)}^n(x-1{)}^n}{n{2}^n}$. Apply the Ratio Test: $$\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\frac{|x-1|}{2}\underset{n\to \infty }{\lim }\frac{n}{n+1}=\frac{|x-1|}{2}$$ For convergence, $\frac{|x-1|}{2}<1⟹|x-1|<2$, so radius of convergence $R=2$.

(b) The open interval is $(1-2,1+2)=(-1,3)$. Test endpoints:

• Left endpoint $x=-1$: substitute to get ${\sum }_{n=1}^{\infty }\frac{(-1{)}^n(-2{)}^n}{n{2}^n}={\sum }_{n=1}^{\infty }\frac{1}{n}$, a divergent p-series ($p=1$). Exclude $x=-1$.
• Right endpoint $x=3$: substitute to get ${\sum }_{n=1}^{\infty }\frac{(-1{)}^n(2{)}^n}{n{2}^n}={\sum }_{n=1}^{\infty }\frac{(-1{)}^n}{n}$, a convergent alternating harmonic series. Include $x=3$. Final interval of convergence: $(-1,3]$.

(c) The series converges absolutely for all $x$ inside the open interval $(-1,3)$, since the Ratio Test guarantees absolute convergence there. At $x=-1$ the series diverges, and at $x=3$ the series converges conditionally, not absolutely. So the series converges absolutely for all $x\in (-1,3)$.

Question 3 (Application / Real-World Style)

The electric potential along the axis of a charged ring of radius $R_0=1$ m is approximated by the power series $V(x)=V_0{\sum }_{n=0}^{\infty }{\left(\frac{x}{R_0}\right)}^{2n}$ for $|x|<R_0$, where $x$ is distance from the center of the ring in meters and $V_0$ is the constant potential at $x=0$. Find the interval of convergence of this power series, and interpret your result in context.

Worked Solution: Rewrite the series in standard form as ${\sum }_{n=0}^{\infty }\frac{1}{R_0^{2n}}{x}^{2n}$, center $a=0$. Apply the Ratio Test: $$\underset{n\to \infty }{\lim }\left|\frac{x^{2n+2}/R_0^{2n+2}}{x^{2n}/R_0^{2n}}\right|=\frac{x^2}{R_0^2}=x^2$$ For convergence, $x^2<1⟹|x|<1$, so $R=1$. Test endpoints $x=1$ and $x=-1$: substitute to get ${\sum }_{n=0}^{\infty }{1}^n={\sum }_{n=0}^{\infty }1$, which diverges by the nth-Term Test. So the interval of convergence is $(-1,1)$. Interpretation: The power series approximation for the electric potential is only valid for points within 1 meter of the center of the ring, and diverges for points 1 meter or farther from the center.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundational prerequisite for all work with Taylor and Maclaurin series, which make up the majority of the series unit on the AP Calculus BC exam. Without correctly finding the interval of convergence for a Taylor series, you cannot use that series to approximate functions, derivatives, or definite integrals, which is a common high-weight FRQ task. Mastering convergence testing also prepares you to distinguish between absolute and conditional convergence, a key skill for justifying your answers on the exam. After this topic, you will move on to constructing Taylor polynomials and Taylor series, then using those series for approximations and error bounding.

• Taylor and Maclaurin polynomials and series
• Absolute and conditional convergence of series
• Approximation with Taylor series
• Error bounds for Taylor polynomials