Radius and Interval of Convergence of Power Series - AP Calculus BC Study Guide

The Core Idea: Radius and Interval of Convergence of Power Series

A power series is an infinite series of the form $\sum a_n(x-c{)}^n$, which can be thought of as a polynomial of infinite degree centered at $x=c$. Unlike a standard polynomial that is defined for all real numbers, a power series may only converge (sum to a finite value) for a specific set of $x$ values. The central problem of this topic is to determine this exact set of values.

This set of convergence is always an interval, known as the interval of convergence. The radius of convergence, $R$, describes the "size" of this interval. For a series centered at $c$, the series will converge for all $x$ values within a distance $R$ from the center ($|x-c|<R$) and diverge for all $x$ values farther than $R$ from the center ($|x-c|>R$). The behavior at the endpoints of the interval, $x=c-R$ and $x=c+R$, must be investigated separately.

Key Formulas/Rules/Theorems

The primary tool for determining the radius and interval of convergence is the Ratio Test.

The Ratio Test for Power Series

For a power series $\sum a_n(x-c{)}^n$, we compute the limit:

$$L=\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}(x-c{)}^{n+1}}{a_n(x-c{)}^n}\right|$$

The convergence of the series depends on the value of $L$:

1. If $L<1$, the series converges absolutely. This inequality is solved for $x$ to find the open interval of convergence.

2. If $L>1$, the series diverges.

3. If $L=1$, the Ratio Test is inconclusive. This situation occurs at the endpoints of the interval of convergence, requiring the use of other convergence tests.

From the condition for convergence, $L<1$, we derive an inequality of the form $|x-c|<R$.

• $c$ is the center of the series.

• $R$ is the radius of convergence.

Understanding the Radius and Interval

The process of finding the interval of convergence is a systematic application of the Ratio Test followed by an analysis of the endpoints.

1. Finding the Radius ($R$): After applying the Ratio Test and simplifying, you will arrive at an expression like $L=k\cdot |x-c|$, where $k$ is a constant derived from the coefficients $a_n$. Setting $L<1$ gives $k\cdot |x-c|<1$, which simplifies to $|x-c|<1/k$. The radius of convergence is therefore $R=1/k$.

2. The Three Cases for $R$:

   • Finite Radius ($0<R<\infty$): The series converges on the open interval $(c-R,c+R)$. You must then test the two endpoints, $x=c-R$ and $x=c+R$, by substituting them back into the original series and using other convergence tests (e.g., p-Series Test, Alternating Series Test, etc.).

   • Zero Radius ($R=0$): This occurs if the limit $L$ is infinite for any $x\ne c$. In this case, the series converges only at its center, $x=c$. The interval of convergence is the single point \{c\}Formula[1]L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(x-3)^{n+1}}{(n+1) \cdot 2^{n+1}} \cdot \frac{n \cdot 2^n}{(x-3)^n} \right|Formula[2]L = \lim_{n \to \infty} \left| \frac{(x-3) \cdot n \cdot 2^n}{(n+1) \cdot 2 \cdot 2^n} \right| = \lim_{n \to \infty} \left| \frac{x-3}{2} \cdot \frac{n}{n+1} \right|Formula[3]L = \left| \frac{x-3}{2} \right| \lim_{n \to \infty} \left( \frac{n}{n+1} \right)Formula[4]L = \left| \frac{x-3}{2} \right|Formula[5]\left| \frac{x-3}{2} \right| < 1 \implies |x-3| < 2Formula[6]\sum_{n=1}^{\infty} \frac{(1-3)^n}{n \cdot 2^n} = \sum_{n=1}^{\infty} \frac{(-2)^n}{n \cdot 2^n} = \sum_{n=1}^{\infty} \frac{(-1)^n 2^n}{n \cdot 2^n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n}Formula[7]\sum_{n=1}^{\infty} \frac{(5-3)^n}{n \cdot 2^n} = \sum_{n=1}^{\infty} \frac{2^n}{n \cdot 2^n} = \sum_{n=1}^{\infty} \frac{1}{n}Formula[8]L = \lim_{n \to \infty} \left| \frac{(n+2)x^{2n+2}}{9^{n+1}} \cdot \frac{9^n}{(n+1)x^{2n}} \right|Formula[9]L = \lim_{n \to \infty} \left| \frac{(n+2)}{(n+1)} \cdot \frac{9^n}{9^{n+1}} \cdot \frac{x^{2n+2}}{x^{2n}} \right| = \lim_{n \to \infty} \left| \frac{n+2}{n+1} \cdot \frac{1}{9} \cdot x^2 \right|Formula[10]L = \frac{|x^2|}{9} \lim_{n \to \infty} \left( \frac{n+2}{n+1} \right)Formula[11]L = \frac{x^2}{9}Formula[12]\frac{x^2}{9} < 1 \implies x^2 < 9 \implies |x| < 3Formula[13]\sum_{n=0}^{\infty} \frac{(n+1)(-3)^{2n}}{9^n} = \sum_{n=0}^{\infty} \frac{(n+1)(9^n)}{9^n} = \sum_{n=0}^{\infty} (n+1)Formula[14]\sum_{n=0}^{\infty} \frac{(n+1)(3)^{2n}}{9^n} = \sum_{n=0}^{\infty} \frac{(n+1)(9^n)}{9^n} = \sum_{n=0}^{\infty} (n+1)Formula[15] and stop, failing to earn points for checking the behavior at $x=a and $x=b$.

• Algebraic Errors in the Ratio Test: Simple mistakes when simplifying the fraction $\frac{a_{n+1}}{a_n}$, especially with factorials (e.g., $\frac{n!}{(n+1)!}=\frac{1}{n+1}$) or powers.

• Incorrect Endpoint Analysis: Choosing an inappropriate test for the endpoint series or misinterpreting the result of a test. For example, concluding that $\sum \frac{1}{n^2}$ diverges.

• Absolute Value Errors: Incorrectly solving the inequality $|x-c|<R$. Forgetting that this is equivalent to $-R<x-c<R$.

• Confusing Radius and Interval: Stating the radius is an interval (e.g., $R=(-2,2)$) or the interval is a single number (e.g., Interval = 4). The radius $R$ is a non-negative number; the interval is a set of $x$-values.