Radius and Interval of Convergence of Power Series

Imagine building a math formula that works well near one point, then starts to fail as you move too far away. That is exactly what happens with many power series 📈. In this lesson, students, you will learn how to find where a power series actually works, how far it works, and what happens at the edges.

By the end of this lesson, you should be able to:

• explain what a power series is and what convergence means,
• find the radius and interval of convergence of a power series,
• test the endpoints carefully,
• connect this topic to Taylor and Maclaurin series,
• and use AP Calculus BC reasoning to justify your answers.

This topic matters because power series are one of the main tools for representing functions in AP Calculus BC. They connect limits, sequences, series, and approximation all in one place. 🌟

What Is a Power Series?

A power series is an infinite series written in the form $\sum_{n=0}^{\infty} a_n (x-c)^n$ or sometimes $\sum_{n=1}^{\infty} a_n (x-c)^n$, where $c$ is the center of the series and $a_n$ are constants.

Think of $c$ as the “home base” of the series. The value of $x$ tells us how far we are from that home base. If $x$ stays close enough to $c$, the series may converge. If $x$ is too far away, the series may diverge.

A key idea is that a power series does not always work for every real number. Instead, it usually converges on some interval around $c$. That interval is the focus of this lesson.

For example, consider $\sum_{n=0}^{\infty} x^n$. This is a power series centered at $0$. It converges when $|x|<1$ and diverges when $|x|\ge 1$. So its behavior depends completely on the input value of $x$.

Radius of Convergence: How Far the Series Works

The radius of convergence is the distance from the center $c$ to the boundary where the series changes from convergence to divergence.

If a power series converges for all $x$ such that $|x-c|<R,$ then $R$ is called the radius of convergence.

There are three possible cases:

• $R$ is a positive number, so the series converges inside an interval around $c$.
• $R=0$, so the series converges only at $x=c$.
• $R=\infty$, so the series converges for every real number.

A common AP Calculus BC method for finding $R$ is the Ratio Test. Suppose we have a power series $\sum a_n (x-c)^n$. If the Ratio Test gives a condition like $|x-c|<R,$ then that inequality tells us the radius immediately.

Example 1

Find the radius of convergence for $$\sum_{n=1}^{\infty} \frac{(x-2)^n}{n}.$$

Using the Ratio Test on the general term $\frac{(x-2)^n}{n},$ we compare consecutive terms and get a condition equivalent to $$|x-2|<1.$$

So the radius of convergence is $$R=1.$$

That means the series converges for values of $x$ within 1 unit of $2$. But that is not the full answer yet. We still need the endpoints.

Interval of Convergence: The Full Picture

The interval of convergence is the set of all $x$ values for which the series converges. It always includes the inside points where $|x-c|<R,$ and it may or may not include the endpoints.

If the radius is $R$, then the basic interval is

$$c-R < x < c+R.$$

But the final interval can change after testing endpoints.

Why do endpoints need separate testing? Because the tests used to find the radius usually tell us what happens when $x$ is strictly inside the interval, but they do not always decide what happens when $x=c-R$ or $$x=c+R.$$

This is a very important AP Calculus BC idea: the endpoint cases can behave differently, even when the middle of the interval is already known.

Example 2

Return to $\sum_{n=1}^{\infty} \frac{(x-2)^n}{n}.$ We already found $R=1,$ so we test the endpoints $x=1$ and $x=3$.

At $x=1$, the series becomes

$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n}.$$

This is the alternating harmonic series, which converges.

At $x=3$, the series becomes

$$\sum_{n=1}^{\infty} \frac{1}{n},$$

which is the harmonic series and diverges.

So the interval of convergence is

$$[1,3).$$

Notice how one endpoint is included and the other is not. That is normal.

How to Find the Radius and Interval Efficiently

The most common AP strategy is:

1. Write the general term of the power series.
2. Apply the Ratio Test or sometimes the Root Test.
3. Solve the inequality for $x$.
4. Get the radius $R$ from the inequality.
5. Test both endpoints separately.
6. Write the final interval of convergence.

Let’s look at another example.

Example 3

Find the interval of convergence for $$\sum_{n=0}^{\infty} \frac{x^n}{2^n}.$$

This can be rewritten as $\sum_{n=0}^{\infty} \left(\frac{x}{2}\right)^n,$ which is geometric.

A geometric series converges when $\left|\frac{x}{2}\right|<1,$ so $|x|<2.$ Therefore the radius of convergence is $$R=2.$$

Now test the endpoints:

• At $x=2$, the series is $\sum_{n=0}^{\infty} 1,$ which diverges.
• At $x=-2$, the series is $\sum_{n=0}^{\infty} (-1)^n,$ which also diverges because the terms do not approach $0$.

So the interval of convergence is

$$(-2,2).$$

What the Endpoints Can Do

At the endpoints, several different things can happen. This is why the interval of convergence must be checked carefully.

An endpoint series might:

• converge absolutely,
• converge conditionally,
• or diverge.

For AP Calculus BC, you do not need a separate “endpoint formula.” You need to test each endpoint using a familiar convergence test.

Common endpoint tools include:

• the Divergence Test,
• the p-series test,
• the Alternating Series Test,
• comparison or limit comparison,
• or recognizing a geometric or harmonic series.

Example 4

Consider $$\sum_{n=1}^{\infty} \frac{(x-4)^n}{n^2}.$$

Using the Ratio Test, the power of $(x-4)^n$ gives the condition $|x-4|<1,$ so $$R=1.$$

Now test the endpoints:

• At $x=3$, the series becomes $\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2},$ which converges absolutely because $\sum_{n=1}^{\infty} \frac{1}{n^2}$ is a convergent p-series.
• At $x=5$, the series becomes $\sum_{n=1}^{\infty} \frac{1}{n^2},$ which also converges.

So the interval of convergence is

$$[3,5].$$

This example shows that both endpoints can be included.

Connection to Taylor and Maclaurin Series

Power series are the foundation of Taylor and Maclaurin series. A Taylor series centered at $c$ has the form

$$\sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x-c)^n.$$

A Maclaurin series is just a Taylor series centered at $0$:

$$\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n.$$

These series are power series, so they also have a radius and interval of convergence.

This matters because a Taylor series is only useful where it converges to the function. For example, the Maclaurin series for $\frac{1}{1-x}$ is

$$\sum_{n=0}^{\infty} x^n,$$

which converges only when $|x|<1.$ That means the function is represented by the series only inside that interval.

In AP Calculus BC, this connection helps explain why approximation works well near the center and becomes weaker farther away.

Conclusion

Radius and interval of convergence tell us where a power series works. The radius $R$ gives the distance from the center $c$ to the boundary of convergence, and the interval gives the complete set of $x$ values where the series converges. The inside of the interval comes from the main convergence test, but the endpoints must be checked one by one.

This topic connects sequences, series, convergence tests, and Taylor polynomials in a single framework. When students understands radius and interval of convergence, it becomes much easier to study power series, build approximations, and analyze functions in AP Calculus BC. ✅

Study Notes

• A power series has the form $\sum_{n=0}^{\infty} a_n (x-c)^n$ and is centered at $c$.
• The radius of convergence $R$ is the distance from $c$ to the boundary where convergence stops.
• The interval of convergence is the set of all $x$ values for which the series converges.
• The inside of the interval is usually found from the Ratio Test or Root Test.
• The condition often looks like $$|x-c|<R.$$
• Endpoints must be tested separately.
• Endpoint tests may use the Divergence Test, p-series test, Alternating Series Test, or recognition of geometric series.
• A series may converge on neither, one, or both endpoints.
• A power series can have $R=0$, a positive $R$, or $R=\infty$.
• Taylor and Maclaurin series are special power series, so they also have radii and intervals of convergence.
• In AP Calculus BC, always show the inequality, test endpoints carefully, and write the final interval in interval notation.