Radius and interval of convergence of power series

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

1. Core Definitions of Power Series Convergence★★☆☆☆⏱ 3 min

A power series is an infinite series centered at constant $a$ , with constant coefficients $c_{n}$ . For any input $x$ , the series reduces to a numerical infinite series that either converges or diverges. This topic makes up 17-18% of the total AP Calculus BC exam score, and appears in both multiple-choice and free-response sections, almost always paired with other series topics like Taylor series.

📘 Definition

Power Series

$\sum_{n = 0}^{\infty} c_{n} (x - a)^{n}$

An infinite series centered at constant $a$ , where each term depends on powers of $(x - a)$ and constant coefficients $c_{n}$ .

📘 Definition

Radius of Convergence

$R$

Half the length of the interval of convergence, equal to the maximum distance from the center $a$ where the series converges. If the series converges only at $a$ , $R = 0$ ; if it converges for all real $x$ , $R = \infty$ .

📘 Definition

Interval of Convergence

The set of all real $x$ for which the power series converges, which is always an interval centered at $a$ .

2. Finding Radius of Convergence★★☆☆☆⏱ 4 min

The standard method for finding $R$ on the AP exam is the Ratio Test, which works seamlessly for factorial, polynomial, and exponential terms common in Taylor and Maclaurin series. 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, substituting the general term $a_{n} = c_{n} (x - a)^{n}$ gives:

L = |x - a| \cdot \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right|

Requiring $L < 1$ for convergence gives the radius of convergence formula:

R = \frac{1}{\lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right|}

The Root Test is an alternative for series with terms raised to the nth power: $R = 1/ (lim_{n \to \infty} n ∣ c_{n} ∣)$ , but this is rarely tested on the AP exam. 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:
3
$\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{3^{n+1} (x-4)^{n+1}}{(n+1)!} \cdot \frac{n!}{3^n (x-4)^n} \right| = \frac{3 |x-4|}{n+1}$
4
Evaluate the limit as $n \to \infty$ :
5
$L = \lim_{n \to \infty} \frac{3 |x-4|}{n+1} = 0 \text{ for any finite } x$
6
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★★★☆☆⏱ 5 min

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. 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. The AP exam explicitly tests whether you remember to check endpoints—omitting this step costs points on FRQ.

📐 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:
3
$\sum_{n=1}^{\infty} \frac{(-5+2)^n}{n \cdot 3^n} = \sum_{n=1}^{\infty} \frac{(-3)^n}{n \cdot 3^n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n}$
4
This is the alternating harmonic series, which converges by the Alternating Series Test, so include $x = - 5$ .
5
Test the right endpoint $x = 1$ : substitute $x = 1$ into the series:
6
$\sum_{n=1}^{\infty} \frac{(1+2)^n}{n \cdot 3^n} = \sum_{n=1}^{\infty} \frac{3^n}{n \cdot 3^n} = \sum_{n=1}^{\infty} \frac{1}{n}$
7
This is the harmonic p-series with $p = 1$ , which diverges, so exclude $x = 1$ .
8
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 of Convergence★★★☆☆⏱ 3 min

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 coefficients $c_{n}$ grow so quickly that for any $x \neq = a$ , 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 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 $\sum_{n = 0}^{\infty} n! (4 x - 8)^{n}$ .

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:
3
$\lim_{n \to \infty} \left| \frac{(n+1)! 4^{n+1} (x-2)^{n+1}}{n! 4^n (x-2)^n} \right| = \lim_{n \to \infty} 4(n+1)|x-2|$
4
For any $x \neq = 2$ , this limit is $\infty > 1$ , so the series diverges for all $x \neq = 2$ .
5
At $x = 2$ , the series becomes $\sum_{n = 0}^{\infty} n! (0)^{n} = 1 + 0 + 0 + ... = 1$ , which converges.
6
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. Check Your Understanding★★☆☆☆⏱ 2 min

✓ Quick check

Try this AP-style multiple choice question:

What is the radius of convergence of the power series $\sum_{n = 0}^{\infty} \frac{( 2 x ) ^{n}}{( n + 1 ) ^{3}}$ ?
- $\frac{1}{2}$
- $1$
- $2$
- $\infty$
Reveal answer
$\frac{1}{2}$ —
Correct: Rewrite the series as $\sum \frac{2 ^{n}}{( n + 1 ) ^{3}} x^{n}$ , apply the Ratio Test to get $∣ x ∣ < 1/2$ , so $R = 1/2$ .

6. Common Pitfalls

Wrong move:

After finding $R$ , stopping and writing the interval as $(a - R, a + R)$ without testing endpoints

Why:

AP exam problems are designed to have one or both endpoints converge, explicitly testing this step, and omitting it loses points

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 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 series is given as $\sum c_{n} (k x - a)^{n}$ , failing to factor out $k$ before calculating $R$ , leading to an incorrect interval

Why:

Students treat $k x - a$ 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 $R$

Wrong move:

When $L = 1$ at an endpoint, concluding 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

7. Quick Reference Cheatsheet

Category	Formula / Result	Notes
General Power Series Form	$\sum_{n = 0}^{\infty} c_{n} (x - a)^{n}$	$a$ = center, $c_{n}$ = constant coefficients
Radius of Convergence (Ratio Test)	$R = \frac{1}{l i m _{n \to \infty} ∣ \frac{c _{n + 1}}{c _{n}} ∣}$	Use for factorials/polynomials; 90% of AP problems
Radius of Convergence (Root Test)	$R = \frac{1}{l i m _{n \to \infty} n ∣ c _{n} ∣}$	Use for terms raised to the nth power
Open Interval of Convergence	$(a - R, a + R)$	Absolute convergence guaranteed here
Convergence only at center	$R = 0$ , Interval = ${a}$	Diverges for all $x \neq = a$
Convergence for all real $x$	$R = \infty$ , Interval = $(- \infty, \infty)$	Limit $L = 0 < 1$ for all finite $x$
Endpoint Testing Rule	Test $x = a - R$ and $x = a + R$	Use non-Ratio tests: AST for alternating, p-Series for positive
Non-standard Form Adjustment	$\sum c_{n} (k x - a)^{n} = \sum c_{n} k^{n} (x - \frac{a}{k})^{n}$	Factor out coefficient of $x$ first

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 · MCQ
Find radius of convergence
2022 · FRQ
Find full interval of convergence
2021 · MCQ
Identify correct interval endpoints

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 free-response question 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 mastering this topic, you will move on to constructing Taylor polynomials and Taylor series, then using those series for approximations and error bounding.