01 / 21
R

Power Series
Radius & Interval of Convergence

Master the techniques for finding where power series converge, from the Ratio Test to checking endpoints.

AP Calculus BC

What Is a Power Series?

A power series centered at \( c \) is an infinite series of the form:

Each term contains a coefficient \( a_n \) and a power of \( (x - c) \). The series behaves like a polynomial of infinite degree that may converge only for certain values of \( x \).

Convergence Behavior

A power series \( \sum a_n(x-c)^n \) has exactly three possible convergence scenarios:

Case 1
Converges only at \( x = c \). The radius is \( R = 0 \).
Case 2
Converges for all \( x \) with \( |x - c| < R \) and diverges for \( |x - c| > R \).
Case 3
Converges for all real \( x \). The radius is \( R = \infty \).

Visualizing the Interval

\(c\) \(c - R\) \(c + R\) Converges Diverges Diverges \(R\)

Radius vs. Interval

Radius of Convergence \( R \)
A single non-negative number (or \( \infty \)) that measures how far from the center \( c \) the series converges. Found via the Ratio or Root Test.
Interval of Convergence (IOC)
The set of all \( x \)-values where the series converges. Includes endpoint analysis: could be \( (c-R,c+R) \), \( [c-R,c+R) \), \( (c-R,c+R] \), or \( [c-R,c+R] \).

The Ratio Test for Power Series

For \( \sum a_n (x - c)^n \), compute:

The series converges when \( |x - c| \cdot L < 1 \), i.e., when \( |x - c| < \frac{1}{L} = R \). If \( L = 0 \), then \( R = \infty \). If \( L = \infty \), then \( R = 0 \).

The Root Test Alternative

The Root Test is especially useful when \( a_n \) involves nth powers (e.g., \( a_n = \left(\frac{n}{2n+1}\right)^n \)).

When to use Ratio
Factorials, products, or ratios in \( a_n \) (e.g., \( \frac{n!}{3^n} \)).
When to use Root
Expressions raised to the \( n \)th power (e.g., \( \left(\frac{2}{3}\right)^n \)).

Open vs. Closed Endpoints

The Ratio/Root Test is inconclusive at \( |x - c| = R \). You must substitute each endpoint into the original series and test with a separate convergence test:

Converges at endpoint
Use a bracket [ or ]. Common tests: Alternating Series Test, p-series comparison.
Diverges at endpoint
Use a parenthesis ( or ). Often the resulting series is harmonic or grows without bound.

Step-by-Step Method

1
Identify \( a_n \) and the center \( c \) from the series \( \sum a_n(x-c)^n \).
2
Apply the Ratio Test: compute \( L = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| \).
3
Find R: set \( R = \frac{1}{L} \). The open interval is \( (c - R,\ c + R) \).
4
Test endpoints: substitute \( x = c - R \) and \( x = c + R \) into the series. Use appropriate tests (AST, p-series, etc.).
5
Write the IOC using brackets [ ] for convergent endpoints and parentheses ( ) for divergent ones.

Example 1: \( \sum \frac{x^n}{n} \)

Find the radius and interval of convergence for \( \displaystyle\sum_{n=1}^{\infty} \frac{x^n}{n} \). Here \( c = 0 \) and \( a_n = \frac{1}{n} \).

Example 2: \( \sum \frac{(x-3)^n}{n \cdot 2^n} \)

Find the radius and interval of convergence. Here \( c = 3 \) and \( a_n = \frac{1}{n \cdot 2^n} \).

Example 3: \( \sum \frac{n!\, x^n}{n^n} \)

Find \( R \) for \( \displaystyle\sum_{n=1}^{\infty} \frac{n!\, x^n}{n^n} \). Factorials call for the Ratio Test.

The open interval is \( (-e,\, e) \). Endpoint tests would require Stirling's approximation: a more advanced topic.

Special Case: Geometric Series

The geometric series \( \sum_{n=0}^{\infty} x^n = \frac{1}{1-x} \) is the simplest power series. It converges when \( |x| < 1 \), so \( R = 1 \).

x y -1 1 R = 1 0 \(\frac{1}{1-x}\)

Endpoint Test Toolkit

After finding \( R \), substitute each endpoint and recognize the resulting series:

1
p-series: \( \sum \frac{1}{n^p} \) converges if \( p > 1 \), diverges if \( p \leq 1 \).
2
Alternating Series Test: if \( |b_n| \) decreases to 0, then \( \sum (-1)^n b_n \) converges.
3
Divergence Test: if \( \lim_{n\to\infty} a_n \neq 0 \), the series diverges.
4
Comparison/Limit Comparison: compare to a known series to settle convergence.

Common Mistakes

✕ Forgetting Endpoints
Finding \( R \) but not testing \( x = c \pm R \). The Ratio Test is inconclusive at these values.
✕ Wrong Center
Using \( |x| < R \) when the series is centered at \( c \neq 0 \). It should be \( |x - c| < R \).
✕ Confusing R and IOC
Writing "the radius is \( [-1, 3] \)" or "the interval is 2". \( R \) is a number; the IOC is an interval.

Summary Comparison

Series Center R IOC
\( \sum \frac{x^n}{n} \) 0 1 \( [-1,1) \)
\( \sum \frac{(x-3)^n}{n \cdot 2^n} \) 3 2 \( [1,5) \)
\( \sum x^n \) 0 1 \( (-1,1) \)
\( \sum \frac{x^n}{n!} \) 0 \( \infty \) \( (-\infty,\infty) \)
\( \sum n!\, x^n \) 0 0 \( \{0\} \)

Question 1 of 5

What is the radius of convergence of \( \displaystyle\sum_{n=0}^{\infty} \frac{x^n}{3^n} \)?

Question 2 of 5

For \( \displaystyle\sum_{n=1}^{\infty} \frac{(x+2)^n}{n^2} \), what is the interval of convergence?

Question 3 of 5

The series \( \displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!} \) has radius of convergence:

Question 4 of 5

If a power series has \( R = 4 \) and is centered at \( c = 5 \), which value of \( x \) is guaranteed to be in the interval of convergence?

Question 5 of 5

At the endpoint of its interval, \( \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{n} \) converges. Which test best justifies this?