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:

$$L=\underset{n\to \infty }{\lim }\left|\frac{c_{n+1}}{c_n}\right|\cdot |x-a|$$ 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 ($L<1$):

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

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:

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

1. Set $L<1$ for convergence:

$$\frac{|x-2|}{5}<1\Rightarrow |x-2|<5$$ 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 $-3<x\le 7$. 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 $R$ ($0\le R\le \infty$) 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 $R=\frac{1}{{\lim }_{n\to \infty }|\frac{c_{n+1}}{c_n}|}$ 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.