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Reading Time: 5 min
Last Updated: March 30, 2026
Main Ideas: 5
Reading Time: 5 min
Last Updated: March 30, 2026
Main Ideas: 5

Topic 10.13 Notes – Radius and Interval of Convergence of Power Series

Verified for 2027 AP® Calculus BC Exam
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Given an infinite “polynomial-like” expression, you’ll find the radius of convergence and the full interval of convergence. This tells you exactly which xx-values make the series behave nicely and represent a real function.

1. What a Power Series Is and What Convergence Means

A power series has the form

∑n=0∞an(x−c)n \sum_{n=0}^{\infty} a_n (x - c)^n

  • ana_n = coefficients (a sequence of numbers)
  • cc = center
  • xx = variable

You can think of it as an infinite polynomial centered at x=cx = c.

How power series behave

If a power series converges, only two things can happen:

  • It converges only at one point (just x=cx = c), or
  • It converges on an interval centered at cc.

That interval always looks like:

∣x−c∣<R |x - c| < R

where RR is the radius of convergence.

Inside that open interval:

  • The series converges absolutely.
  • It acts like a smooth, well-behaved function.
  • If R>0R > 0, the series is the Taylor series of the function it equals on that interval.

Here’s the geometric picture you should have in your head. The three number lines show the only possible behaviors.

Study guide illustration

Possible convergence behaviors of a power series

The radius gives you the open interval. The endpoints require extra work.

2. Using the Ratio Test to Find the Radius of Convergence

For almost every power series problem, the Ratio Test is the main tool.

Given
∑an(x−c)n \sum a_n (x-c)^n

compute

L=lim⁡n→∞∣an+1(x−c)n+1an(x−c)n∣ L = \lim_{n\to\infty} \left| \frac{a_{n+1}(x-c)^{n+1}}{a_n (x-c)^n} \right|

What actually happens algebraically

After canceling powers, you’ll always end up with:

L=∣x−c∣⋅(limit involving n) L = |x-c| \cdot (\text{limit involving } n)

Call that constant kk. Then:

L=k∣x−c∣ L = k|x-c|

For convergence, the Ratio Test requires L<1L < 1. So:

k∣x−c∣<1 k|x-c| < 1

Solve for ∣x−c∣|x-c|:

∣x−c∣<1k |x-c| < \frac{1}{k}

So the radius is
R=1k R = \frac{1}{k}

Special cases

  • If the limit becomes 0 → converges for all xx, so R=∞R = \infty.
  • If the limit forces x=cx=c only → R=0R = 0.

Common algebra mistake: forgetting to factor out ∣x−c∣|x-c| before taking the limit. Keep it separate from the nn-stuff.

3. Finding the Interval of Convergence

The radius gives you the open interval:

∣x−c∣<R⇒c−R<x<c+R |x-c| < R \quad \Rightarrow \quad c - R < x < c + R

But you are not finished.

You must test:

  • x=c−Rx = c - R
  • x=c+Rx = c + R

When you plug those in, the Ratio Test will give L=1L = 1. It always fails at endpoints. So now you switch to:

  • p-series test
  • Alternating Series Test
  • Comparison
  • Harmonic recognition
  • etc.

Each endpoint is tested independently.

Your final answer should clearly state:

  • Center: cc
  • Radius: RR
  • Interval: use parentheses or brackets correctly

Example structure:
[c−R, c+R) [c-R,\, c+R)

On AP FRQs, you lose points if you forget to test endpoints or if you give only RR when the interval is requested.

4. What the Radius Tells You About the Function

If R>0R > 0, then on (c−R,c+R)(c-R, c+R):

  • The power series equals its function.
  • You can differentiate term-by-term.
  • You can integrate term-by-term.

Very important fact:

  • Differentiating does not change RR.
  • Integrating does not change RR.

The endpoints might behave differently after differentiation or integration, but the radius stays the same. That’s because RR depends on long-term growth of coefficients, and differentiation/integration doesn’t change that growth pattern.

5. Common Exam Mistakes

  • Leaving (x−c)n(x-c)^n out of the Ratio Test.
  • Solving k∣x−c∣<1k|x-c|<1 incorrectly.
  • Forgetting absolute value when solving.
  • Not testing endpoints.
  • Using the Ratio Test at endpoints.
  • Mixing up center cc and radius RR.

If you stay systematic, these problems are very predictable.

Key Takeaways

A power series either converges at one point or on an interval centered at cc.
The Ratio Test almost always gives RR through an inequality of the form k∣x−c∣<1k|x-c|<1.
The radius gives the open interval; endpoints must be tested separately.
The Ratio Test is inconclusive at endpoints because it gives L=1L=1.
If R>0R>0, the series is the Taylor series of its function on (c−R,c+R)(c-R, c+R).
Term-by-term differentiation and integration keep the same radius of convergence.

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Notes

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