Radius and Interval of Convergence

2 min read

Reference: Stewart §11.8

Chapter 11.8: Power Series | Course: MATH163 | Difficulty: Intermediate | Time: ~20 min

Textbook Reference


Primary source	OpenStax Calculus Volume 2, Section 6.1: “Power Series and Functions”
Direct link	https://openstax.org/books/calculus-volume-2/pages/6-1-power-series-and-functions
Textbook used in class	Stewart, Calculus, Section 11.8: “Power Series” (Examples 3, 4, 5)

Opening Scenario

Finding the radius of convergence is one step. The full interval of convergence requires checking the endpoints, and each endpoint is an ordinary constant-term series that requires its own test. This lesson organizes the complete procedure.

Quick Reference

Four-step procedure for $\sum c_n(x-a)^n$:

Apply the Ratio Test to find $L$ in terms of $|x - a|$. Set $L < 1$ and solve for the open interval.
Read off $R$ (the half-width of that interval).
Test $x = a + R$ (right endpoint): plug in, get a numerical series, apply any test.
Test $x = a - R$ (left endpoint): plug in, get a numerical series, apply any test.
Write the interval with the correct brackets.

Key Concepts

1. The Full Endpoint Procedure

Example 1. Find the interval of convergence of $\displaystyle\sum_{n=0}^{\infty} \frac{(x+2)^n}{3^n \sqrt{n+1}}$. (Stewart 11.8, Example 5 type.)

Step 1 (Ratio Test): \[ \left|\frac{a_{n+1}}{a_n}\right| = \frac{|x+2|}{3} \cdot \sqrt{\frac{n+1}{n+2}} \to \frac{|x+2|}{3}. \] Converges when $|x+2|/3 < 1$, i.e., $|x+2| < 3$.

Step 2: $R = 3$, center $a = -2$. Open interval: $(-5, 1)$.

Step 3 (right endpoint $x = 1$): $\sum (3)^n/(3^n\sqrt{n+1}) = \sum 1/\sqrt{n+1}$. $p$-series with $p = 1/2 \leq 1$: diverges.

Step 4 (left endpoint $x = -5$): $\sum (-3)^n/(3^n\sqrt{n+1}) = \sum (-1)^n/\sqrt{n+1}$. Alternating Series Test: $1/\sqrt{n+1} \to 0$ and is decreasing: converges.

Boxed answer: Interval of convergence $[-5, 1)$.

2. When Scaling Changes Things

Substituting $u = cx$ into a known power series produces a new power series with a different interval. The center stays at $0$ but the radius scales by $1/|c|$.

Example 2. Given that $\sum_{n=0}^\infty x^n = 1/(1-x)$ for $|x| < 1$, find the interval of convergence for $\sum_{n=0}^\infty (x/3)^n$.

Substitute $u = x/3$: converges for $|x/3| < 1$, i.e., $|x| < 3$. The series sums to $1/(1 - x/3) = 3/(3-x)$.

Boxed answer: Interval $(-3, 3)$.

3. Summary of Endpoint Behavior

At each endpoint you substitute a specific number and get a standard numerical series. The most common results:

If you get $\sum c/n^p$ with $p > 1$: converges.
If you get $\sum c/n$ (harmonic): diverges.
If you get $\sum (-1)^n/n^p$ with $p > 0$: converges (AST).
If you get a geometric series with $|r| \geq 1$: diverges.

Common misconception

always including both endpoints or always excluding both. The two endpoints are independent. The interval might be open at both, closed at both, or half-open. Each endpoint requires its own test. There is no shortcut that applies uniformly to both.

Common Errors Summary

Error	Correction
Using the Ratio Test limit to decide endpoints	At endpoints $L = 1$, which is inconclusive; use a separate test
Forgetting the center when writing the interval	Interval is $(a-R, a+R)$, not $(-R, R)$, unless $a = 0$
Including an endpoint that gives a divergent series	Check each endpoint independently

Leveled Practice

Problem 1. Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n x^n}{n \cdot 4^n}$.

Show answer

$|a_{n+1}/a_n| = \frac{n}{n+1}\cdot\frac{|x|}{4} \to |x|/4$. Converges for $|x| < 4$: $R = 4$.

At $x = 4$: $\sum (-1)^n/n$: converges (AST). At $x = -4$: $\sum 1/n$: diverges.

Interval: $(-4, 4]$.

Mastery Checklist

Execute the four-step procedure correctly for a shifted power series
Check each endpoint with an appropriate test
Write the interval with correct open/closed bracket notation
Adjust the interval when substitution scales the variable

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