10.13 Radius and Interval of Convergence of Power Series

A power series converges on an interval centered at its center point, often written as $r$, and the Ratio Test helps find the radius of convergence $R$. After you find $R$, you get an open interval, then test both endpoints separately to decide whether the interval is open, closed, or half-open. For AP Calculus BC, show the Ratio Test setup and endpoint tests clearly because endpoint behavior can differ.

How Do You Find Radius and Interval of Convergence?

To find the radius and interval of convergence, apply the Ratio Test to the power series and solve the inequality that makes the limit less than 1. That gives the radius $R$ and the open interval centered at $r$.

Then test both endpoints separately in the original series. The Ratio Test is inconclusive at the endpoints, so the final interval can be open, closed, or half-open depending on whether each endpoint series converges.

Why This Matters for the AP Calculus Exam

Radius and interval of convergence is a BC-only topic, and it pulls together almost every convergence test you have learned in this unit. When a problem hands you a power series, you need to find where it converges before you can use it to approximate a function. On the exam you may need to set up the Ratio Test, solve an absolute value inequality for the radius, and then run separate endpoint checks using tools like the alternating series test, the harmonic series, or p-series reasoning. This rewards clean setup and careful justification, which matters for both multiple-choice and free-response work.

Key Takeaways

• A power series has the form $\sum_{n=0}^{\infty} a_n(x-r)^n$, where r is the center and $a_n$ is a sequence of real numbers.
• The Ratio Test gives you the radius of convergence R from the inequality $|x-r| < R$.
• A converging power series either converges at the single point x = r or on a full interval centered at r.
• The radius gives you an open interval, but you must test both endpoints separately to find the complete interval of convergence.
• Endpoint behavior can differ: one endpoint may converge while the other diverges, so check each one with the right test.
• Term-by-term differentiation or integration keeps the same radius of convergence as the original series.

What's a Power Series?

A power series is a series of the form

$$\sum_{n=0}^{\infty} a_n(x-r)^n$$

where $n$ is a non-negative integer, $a_n$ is a sequence of real numbers, and $r$ is a real number.

Here $a_n$ can be any sequence, most of which you have already seen in earlier topics. The value $r$ tells you where the power series is centered. For example, a power series centered at $x = 3$ uses $(x - 3)$ inside each term. A series centered at $r = 0$ is a Maclaurin series.

Radius and Interval of Convergence

A key question about power series is where the series actually converges. For a given series, you find the radius of convergence R first, then the full interval of convergence.

If a power series converges, it either converges at a single point or has an interval of convergence. The Ratio Test is the standard tool for finding R.

For a power series centered at $x = r$, the one point where you are guaranteed convergence is $x = r$ itself. The Ratio Test lets you expand that to a range of x values. If the radius is positive, the power series is the Taylor series of the function it converges to on that open interval.

Recap: The Ratio Test

For a series $\sum a_n$, let $L=|\lim\limits_{n\to \infty}\frac{a_{n+1}}{a_n}|$.

• If $L < 1$, the series converges.
• If $L > 1$, the series diverges.
• If $L = 1$, the test is inconclusive.

To review the Ratio Test in depth, check out 10.8 Ratio Test for Convergence.

How to Use This on the AP Calculus Exam

Worked Example

Find the interval, radius of convergence, and center for the series:

$$\sum^{\infty}_{n=1}\frac{2^n}{n}(4x-8)^n$$

First, apply the Ratio Test and identify $a_n$ and $a_{n+1}$:

$$a_{n}=\frac{2^{n}}{n}(4x-8)^{n}$$ $$a_{n+1}=\frac{2^{n+1}}{n+1}(4x-8)^{n+1}=\frac{2^n*2}{n+1}(4x-8)^n(4x-8)$$

Then find $L$:

$$L=\lim\limits_{n\to \infty}|\frac{a_{n+1}}{a_n}|=\lim\limits_{n\to \infty}|\frac{\frac{2^n*2}{n+1}(4x-8)^n(4x-8)}{\frac{2^{n}}{n}(4x-8)^{n}}|$$

This looks messy, but $2^n$ and $(4x-8)^n$ cancel, leaving:

$$L=\lim\limits_{n\to \infty}|\frac{2n}{n+1}(4x-8)|$$

Since $(4x - 8)$ does not depend on $n$, factor it out of the limit. Evaluating the remaining limit gives 2:

$$L=(4x-8)*\lim\limits_{n\to \infty}|\frac{2n}{n+1}|=(4x-8)*2$$

The series converges when $L < 1$:

$$|L|<1$$ $$|2(4x-8)|<1$$ $$|2*4(x-2)|<1$$ $$|x-2|<\frac{1}{8}$$

This gives a radius of convergence of $\frac{1}{8}$ and a center at $x = 2$.

To find the interval, use the absolute value inequality $|x-2|<\frac{1}{8}$:

$$x-2<\frac{1}{8}=x<\frac{1}{8}+\frac{16}{8}=x<\frac{17}{8}$$ $$x-2>-\frac{1}{8}=x>-\frac{1}{8}+\frac{16}{8}=x>\frac{15}{8}$$

Testing the Endpoints

The Ratio Test says nothing about $x = \frac{15}{8}$ and $x = \frac{17}{8}$, so plug each into the original series.

At $x = 15/8$:

$$\sum^{\infty}_{n=1}\frac{2^n}{n}(4(\frac{15}{8})-8)^n$$ $$\sum^{\infty}_{n=1}\frac{2^n}{n}(\frac{15}{2}-8)^n$$ $$\sum^{\infty}_{n=1}\frac{2^n}{n}(\frac{-1}{2})^n$$ $$\sum^{\infty}_{n=1}\frac{2^n}{n}(\frac{(-1)^n}{2^n})$$ $$\sum^{\infty}_{n=1}\frac{(-1)^n}{n}$$

This is the alternating harmonic series, which converges. So $x = 15/8$ is included in the interval.

At $x = 17/8$:

$$\sum^{\infty}_{n=1}\frac{2^n}{n}(4(\frac{17}{8})-8)^n$$ $$\sum^{\infty}_{n=1}\frac{2^n}{n}(\frac{17}{2}-8)^n$$ $$\sum^{\infty}_{n=1}\frac{2^n}{n}(\frac{1}{2})^n$$ $$\sum^{\infty}_{n=1}\frac{2^n}{n}(\frac{1}{2^n})$$ $$\sum^{\infty}_{n=1}\frac{1}{n}$$

This is the harmonic series, which diverges. So $x = 17/8$ is not included.

Putting it together, the interval of convergence is $[\frac{15}{8},\frac{17}{8})$ or $\frac{15}{8}\leq x <\frac{17}{8}$.

Problem Solving Routine

When a problem asks for the radius and interval of convergence, follow this order:

1. Apply the Ratio Test to the general term.
2. Solve the resulting inequality $|x-r| < R$ to get the radius and the open interval.
3. Plug each endpoint back into the original series and test it separately.
   • If the endpoint series converges, that endpoint is included (use a bracket).
   • If the endpoint series diverges, that endpoint is not included (use a parenthesis).
   • Pick the right test for each endpoint using your toolkit from 10.3 through 10.9.

Common Trap

Many students stop after finding the radius and report an open interval. The Ratio Test gives $L = 1$ at both endpoints, which is inconclusive, so you cannot skip the endpoint checks. Show that work clearly, since each endpoint can behave differently.

Common Misconceptions

• The radius of convergence is not the interval of convergence. The radius is a single number R; the interval is the actual set of x values, found only after endpoint testing.
• Endpoints are not automatic. The Ratio Test is inconclusive when $|x-r| = R$, so you must test each endpoint with a separate convergence test.
• Both endpoints do not always behave the same way. One can converge while the other diverges, which gives a half-open interval.
• The center r is where the series is guaranteed to converge, not where it diverges. Every power series converges at least at x = r.
• Differentiating or integrating a power series term by term keeps the same radius of convergence, but the endpoint behavior can change, so recheck endpoints if needed.
• A radius of convergence can be 0, a finite number, or infinite. R = ∞ means the series converges for all real x, like the series for $e^x$.

Related AP Calculus Guides

• Unit 10 Overview: Infinite Series and Sequences
• 10.1 Defining Convergent and Divergent Infinite Series
• 10.3 The nth Term Test for Divergence
• 10.5 Harmonic Series and p-Series
• 10.2 Working with Geometric Series
• 10.4 Integral Test for Convergence