AP Calculus BC: Radius and Interval of Convergence

Understanding where a power series converges is not just an abstract exercise; it’s the crucial step that tells you where this powerful tool for representing functions actually works. Whether you're modeling physical systems in engineering or analyzing functions in pure mathematics, a series that diverges is useless. This process—finding the radius of convergence $R$ and the interval of convergence—transforms an infinite sum from a potential mystery into a reliable, well-defined function on a specific domain.

What is Convergence for a Power Series?

A power series centered at $x = a$ is an infinite series of the form: $$\sum_{n=0}^{\infty} c_n (x - a)^n = c_0 + c_1(x-a) + c_2(x-a)^2 + \cdots$$ where $c_n$ are the coefficients. Unlike a simple polynomial, its behavior depends entirely on the value of $x$ you plug in. For some $x$, the sum approaches a finite number (converges); for others, it grows without bound or oscillates (diverges).

A fundamental theorem guarantees that for any power series, there exists a non-negative number $R$, called the radius of convergence. The series will:

• Converge absolutely for all $x$ satisfying $|x - a| < R$.
• Diverge for all $x$ satisfying $|x - a| > R$.

What happens exactly at the radius, where $|x - a| = R$, requires separate investigation. The set of all $x$-values for which the series converges is its interval of convergence. This interval can be $(a-R, a+R)$, $[a-R, a+R)$, $(a-R, a+R]$, or $[a-R, a+R]$.

Think of the center $a$ as the epicenter of a circle of reliability. The radius $R$ tells you how far out that reliability extends before you enter the zone of divergence. Checking the endpoints is like testing the integrity of the boundary wall itself.

Finding R with the Ratio Test

The most common tool for finding $R$ is the Ratio Test. Given a series $\sum a_n$, you examine the limit $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$. The test states that the series converges absolutely if $L < 1$, diverges if $L > 1$, and is inconclusive if $L = 1$.

For a power series $\sum c_n (x-a)^n$, your term $a_n$ is $c_n (x-a)^n$. Applying the Ratio Test:

1. Compute the absolute ratio:

$$\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{c_{n+1}(x-a)^{n+1}}{c_n (x-a)^n} \right| = \left| \frac{c_{n+1}}{c_n} \right| \cdot |x-a|$$

1. Take the limit:

$$L = \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right| \cdot |x-a|$$ Often, the limit $\lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right|$ is a constant, which we can call $K$. Then $L = K \cdot |x-a|$.

1. Apply the Ratio Test condition for absolute convergence ($L < 1$):

$$K \cdot |x-a| < 1 \quad \text{or} \quad |x-a| < \frac{1}{K}$$ This inequality directly reveals the radius of convergence: $R = \frac{1}{K} = \frac{1}{\lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right|}$.

Worked Example: Find the radius of convergence for $\sum_{n=1}^{\infty} \frac{(-1)^{n+1} (x-2)^n}{n \cdot 5^n}$.

1. Identify $a_n = \frac{(-1)^{n+1} (x-2)^n}{n \cdot 5^n}$.
2. Form the ratio:

$$\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{ \frac{(-1)^{n+2} (x-2)^{n+1}}{(n+1) \cdot 5^{n+1}} }{ \frac{(-1)^{n+1} (x-2)^n}{n \cdot 5^n} } \right| = \left| \frac{n (x-2)}{5(n+1)} \right|$$

1. Take the limit:

$$L = \lim_{n \to \infty} \left| \frac{n (x-2)}{5(n+1)} \right| = \lim_{n \to \infty} \frac{n}{n+1} \cdot \frac{|x-2|}{5} = 1 \cdot \frac{|x-2|}{5} = \frac{|x-2|}{5}$$

1. Set $L < 1$ for convergence:

$$\frac{|x-2|}{5} < 1 \quad \Rightarrow \quad |x-2| < 5$$ Thus, the radius of convergence is $R = 5$.

Investigating the Endpoints

The ratio test only gives the open interval $|x-a| < R$ (here, $|x-2| < 5$, or $-3 < x < 7$). Convergence at the endpoints $x = -3$ and $x = 7$ must be checked individually by substituting them into the original series.

• At $x = 7$: The series becomes $\sum_{n=1}^{\infty} \frac{(-1)^{n+1} (5)^n}{n \cdot 5^n} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$. This is the alternating harmonic series, which converges (by the Alternating Series Test).
• At $x = -3$: The series becomes $\sum_{n=1}^{\infty} \frac{(-1)^{n+1} (-5)^n}{n \cdot 5^n} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} (-1)^n 5^n}{n \cdot 5^n} = \sum_{n=1}^{\infty} \frac{(-1)^{2n+1}}{n} = \sum_{n=1}^{\infty} \frac{-1}{n}$. This is $-1$ times the harmonic series, which diverges.

Therefore, the interval of convergence is $(-3, 7]$, or $-3 < x \leq 7$. Notice it includes the right endpoint but not the left.

The Root Test and Special Cases

An alternative to the Ratio Test is the Root Test, where you compute $L = \lim_{n \to \infty} \sqrt[n]{|a_n|} = \lim_{n \to \infty} |c_n|^{1/n} \cdot |x-a|$. The condition $L < 1$ again leads to an inequality of the form $|x-a| < R$, where $R = \frac{1}{\lim_{n \to \infty} |c_n|^{1/n}}$. The Root Test is often useful when each term involves an $n$th power.

Two special cases arise from the limit $L$:

• If $L = 0$ for all $x$ (this happens if, for example, $c_n = \frac{1}{n!}$), then the radius of convergence is infinite ($R = \infty$). The series converges for all real numbers.
• If $L = \infty$ for any $x \ne a$ (this happens if, for example, $c_n = n!$), then the radius of convergence is zero ($R = 0$). The series converges only at its center $x = a$.

Common Pitfalls

1. Forgetting to Check Endpoints: The most frequent and costly error. The ratio and root tests are inconclusive when $L=1$, which occurs exactly at $|x-a| = R$. You must plug each endpoint value into the original series and use other convergence tests (p-series, alternating series, comparison, etc.) to determine inclusion. A radius of $R=5$ does not automatically mean the interval is $[a-5, a+5]$.

1. Misapplying Absolute Value in the Ratio Test: When forming $\left| \frac{a_{n+1}}{a_n} \right|$, you must take the absolute value of the entire fraction, not just the coefficients. Correctly handle the $(x-a)$ term: $|(x-a)^{n+1} / (x-a)^n| = |x-a|$.

1. Algebraic or Limit Simplification Errors: Mistakes often occur when simplifying the ratio of factorials, powers, or polynomials. For example, $(n+1)! = (n+1) \cdot n!$. Write out steps carefully. Also, correctly evaluate limits: $\lim_{n \to \infty} \frac{n}{n+1} = 1$.

1. Confusing Radius and Interval: The radius $R$ is a non-negative number (or infinity). The interval is the actual set of $x$-values. Always state both in your final answer. A correct answer format is: "Radius $R=5$. The interval of convergence is $(-3, 7]$."

Summary

• Every power series $\sum c_n (x-a)^n$ has a radius of convergence $R$ ($0 \leq R \leq \infty$) defining where it converges absolutely.
• The Ratio Test (or Root Test) is applied to find $R$ by solving $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1$ for $|x-a|$.
• The value of $R$ is given by $R = \frac{1}{\lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right|}$ when this limit exists.
• The interval of convergence is found by separately testing each endpoint $x = a \pm R$ using other convergence tests.
• The final interval may be open, closed, or half-open, and you must report it explicitly alongside the radius.