AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/10/10-13)
Unit 10 · Topic 10.13
BC only10.13 Radius and Interval of Convergence of Power Series
BC only. A power series converges for x-values within some distance R of its center, called the radius of convergence. You usually find R with the ratio test, then check each endpoint separately, since the series might converge at both, one or neither.
Key terms
- power series
- center
- radius of convergence
- interval of convergence
- endpoints
Power series
This whole unit is BC only. A power series centered at x = a is Σ (n = 0 to ∞) cₙ(x − a)ⁿ = c₀ + c₁(x − a) + c₂(x − a)² + … . For each value of x, it's a series of numbers that may converge or diverge. The set of x-values where it converges is the interval of convergence.
There are only three possibilities:
- It converges only at x = a (radius R = 0).
- It converges for all real x (radius R = ∞).
- It converges for |x − a| < R and diverges for |x − a| > R, for some positive number R. At the endpoints, x = a − R and x = a + R, anything can happen.
Finding the radius with the ratio test
- Compute |aₙ₊₁/aₙ| for the whole term, including the x part.
- Take the limit as n → ∞. It will usually look like (something)·|x − a|.
- Set that limit < 1 and solve for |x − a|. The number on the right side is R.
- If the limit is 0 for every x, R = ∞. If it's ∞ for every x ≠ a, R = 0.
Checking the endpoints
At an endpoint, the ratio test gives L = 1, so it can't help. Plug each endpoint into the series and test the resulting series of numbers with another test: p-series, the alternating series test, the nth term test or a comparison.
Write the final interval with brackets for endpoints that converge and parentheses for endpoints that diverge, like [1, 5).
The center sits in the middle of the interval of convergence, and the radius is half its length. If a question tells you a series converges on (−1, 7), its center is 3 and its radius is 4.
Facts you can use
Inside its interval, a power series with a positive radius defines a function, and the series is exactly that function's Taylor series about the center. You can also differentiate or integrate a power series one term at a time. The new series has the same radius, but its endpoints can behave differently (10.15), so check them again.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Full interval of convergence
Find the interval of convergence of Σ (n = 1 to ∞) (x − 3)ⁿ/(n·2ⁿ).
Show the solutionHide the solution
- Step 1: Ratio: |aₙ₊₁/aₙ| = |x − 3|ⁿ⁺¹/((n + 1)2ⁿ⁺¹) · (n·2ⁿ)/|x − 3|ⁿ = |x − 3| · n/(2(n + 1)).
- Step 2: Limit: |x − 3|/2. Set |x − 3|/2 < 1: |x − 3| < 2. R = 2, so the open interval is 1 < x < 5.
- Step 3: Endpoint x = 5: Σ 2ⁿ/(n·2ⁿ) = Σ 1/n, the harmonic series, which diverges.
- Step 4: Endpoint x = 1: Σ (−2)ⁿ/(n·2ⁿ) = Σ (−1)ⁿ/n. It alternates, and 1/n is positive, decreasing and approaches 0, so it converges by the alternating series test.
- Step 5: Include x = 1, exclude x = 5.
Answer: Interval of convergence: [1, 5), radius 2.
- Example 2
Radius 0 and radius ∞
Find the radius of convergence of (a) Σ n! xⁿ and (b) Σ xⁿ/n!, both from n = 0 to ∞.
Show the solutionHide the solution
- Step 1: (a) Ratio: (n + 1)! |x|ⁿ⁺¹/(n! |x|ⁿ) = (n + 1)|x|. For any x ≠ 0, this → ∞ > 1, so the series diverges. It converges only at x = 0. R = 0.
- Step 2: (b) Ratio: |x|/(n + 1) → 0 < 1 for every x. It converges for all x. R = ∞.
Answer: (a) R = 0 (converges only at x = 0) (b) R = ∞ (converges for all x)
- Example 3
Trap: both endpoints converge
Find the interval of convergence of Σ (n = 1 to ∞) (2x)ⁿ/n².
Show the solutionHide the solution
- Step 1: Ratio: |2x| · n²/(n + 1)² → |2x|. Set |2x| < 1: |x| < 1/2. R = 1/2.
- Step 2: x = 1/2: Σ 1/n² converges (p = 2).
- Step 3: x = −1/2: Σ (−1)ⁿ/n² converges (absolutely, since Σ 1/n² converges).
- Step 4: Students often assume endpoints always fail. Here both are included.
Answer: [−1/2, 1/2]
Common mistakes
- Forgetting to test the endpoints, or testing them with the ratio test (which always gives L = 1 there).
- Losing the factor in front of x: for (2x)ⁿ, the radius is 1/2, not 1.
- Writing the radius as the whole interval's length. The radius is half the length.
- Writing the interval centered at 0 when the center is a.
On the exam
- A classic BC free-response part asks for the interval of convergence and wants to see your work. You need the ratio test setup, the inequality, and an endpoint test with justification for each endpoint.
- State the test you used at each endpoint, such as “at x = 1, the series is the alternating harmonic series, which converges by the alternating series test.”
Connected topics
Videos
Check yourself
4 questions on 10.13 Radius and Interval of Convergence of Power Series. Pick an answer to see if you got it, and why.
What is the interval of convergence of the power series Σ (n = 1 to ∞) (3x − 1)ⁿ/n²?
What is the interval of convergence of the power series Σ (n = 1 to ∞) n²(x + 1)ⁿ/5ⁿ?
For which values of x does the power series Σ (n = 0 to ∞) n!(x − 2)ⁿ converge?
The function f is defined by the power series f(x) = Σ (n = 1 to ∞) (x − 3)ⁿ/(n · 2ⁿ) = (x − 3)/2 + (x − 3)²/8 + (x − 3)³/24 + ⋯ for all x for which the series converges.
Described function
What is the interval of convergence of the power series for f?
0 of 4 answered