A power series' radius tells you how far from its centre absolute convergence extends. It does not usually settle the boundary points. This AP Calculus BC guide uses original examples to separate the ratio test, endpoint tests and final interval, without treating a solution checklist as a fixed scoring scheme.
For a real power series ∑ cₙ(x−a)ⁿ, a is the centre. With a finite positive radius R, the series converges absolutely when |x−a|<R and diverges when |x−a|>R. You must investigate x=a−R and x=a+R separately. The interval includes an endpoint only if the series actually converges there.
These ideas are in College Board's current Calculus BC framework, topic 10.13. They are BC content, not an assumption that every AP Calculus question asks for a radius. Read the requested quantity: a radius is a number, while an interval specifies which real x-values work.
For a fixed x, call the series term uₙ and, when the ratios are defined eventually, calculate L=limₙ→∞|uₙ₊₁/uₙ|. If L<1, the series converges absolutely. If L>1, including an infinite limit, it diverges. If L=1, the test is inconclusive. A zero term can make a quotient undefined, so check the centre or other such cases directly rather than dividing by zero.
The ratio test applies to numerical series too. At many power-series endpoints its limit is 1, which is why another test is needed. The absence of x after substitution is not a reason to prohibit the ratio test. Also, terms tending to zero are necessary for convergence, but are not sufficient: the harmonic series is a counterexample.
The examples in this guide are teaching examples, not released AP questions. Let S(x)=∑ₙ₌₁^∞(x−2)ⁿ/(n·3ⁿ). For x≠2, the absolute-value ratio is |x−2|·n/[3(n+1)]. Its limit is |x−2|/3. Thus absolute convergence holds for |x−2|<3, or −1<x<5, and R=3. At the centre x=2 every term is zero, so the series converges there directly.
At x=−1 the series becomes ∑ₙ₌₁^∞(−1)ⁿ/n. The positive magnitudes 1/n decrease to zero, so the alternating series test establishes convergence. The absolute-value series ∑1/n diverges; convergence at this endpoint is conditional.
At x=5 the series becomes the harmonic series ∑ₙ₌₁^∞1/n, which diverges. Here the term limit is zero and the ratio-test limit is 1; neither establishes convergence. The p-series test with p=1 gives the needed conclusion. Include the left endpoint and exclude the right: the interval is [−1,5).
The next original variations have the same centre 2 and radius 3. They differ in their boundary behaviour. Every sum in the table starts at n=1.
| Power series | At x=−1 | At x=5 | Interval |
|---|---|---|---|
| ∑(x−2)ⁿ/(n²·3ⁿ) | ∑(−1)ⁿ/n² converges absolutely. | ∑1/n² converges by the p-series test, p=2. | [−1,5] |
| ∑(x−2)ⁿ/3ⁿ | Terms (−1)ⁿ do not tend to zero; diverges. | Terms are 1; diverges. | (−1,5) |
| ∑(−1)ⁿ(x−2)ⁿ/(n·3ⁿ) | ∑1/n diverges. | ∑(−1)ⁿ/n converges conditionally. | (−1,5] |
For the n² denominator the ratio contains n²/(n+1)², whose limit is 1. For the geometric variation it is simply |x−2|/3. In the sign-changing-coefficient variation, absolute values remove (−1)ⁿ from the ratio, so the radius remains 3. At the left endpoint the two alternating factors multiply to +1; at the right only the coefficient sign remains. A right endpoint is not automatically an all-positive series, and a nonzero centre does not determine which bracket is closed.
For the original series ∑ₙ₌₁^∞(x+1)ⁿ/n!, the ratio at x≠−1 is |x+1|/(n+1), which tends to zero for every fixed real x. At x=−1 all terms are zero. Hence R=∞ and the interval is (−∞,∞); there are no finite boundary points to test.
For ∑ₙ₌₁^∞n!(x−4)ⁿ, the ratio at x≠4 is (n+1)|x−4|, which tends to infinity. The terms do not tend to zero and the series diverges away from 4. At x=4 all terms are zero. Thus R=0 and the convergence set is the singleton {4}. The empty strict inequality |x−4|<0 cannot establish this centre value; direct substitution does.
Term-by-term differentiation and integration preserve the radius of a power series, as topic 10.13 states. The endpoint membership can change. For the worked S(x), differentiation inside −1<x<5 gives S′(x)=∑ₙ₌₁^∞(x−2)ⁿ⁻¹/3ⁿ. At −1 its terms are (−1)ⁿ⁻¹/3, and at 5 they are 1/3. Neither tends to zero, so this derivative series has interval (−1,5).
Integrating S from 2 to x gives ∑ₙ₌₁^∞(x−2)ⁿ⁺¹/[n(n+1)3ⁿ] inside that open interval. At either endpoint the absolute-value terms are 3/[n(n+1)]≤3/n². Comparison with a convergent p-series establishes absolute convergence at both endpoints, so the integrated series has interval [−1,5]. These conclusions concern the resulting series; do not claim termwise differentiation is automatically valid at an original boundary point.
When an interval is requested, show the ratio limit or another valid radius argument, solve the strict inequality, substitute both finite endpoints, justify their convergence or divergence, and state the interval. This is a useful way to organise reasoning. It does not assign one point to each written line or guarantee a particular AP score.
A historical example shows why the actual task matters: 2026 Calculus BC Question 6 supplies a convergence condition and asks about a geometric sum, a derivative series, an approximation bound and a combination of series. Its published scoring guidelines assess those particular parts. A universal eight-point radius-and-endpoints scheme does not describe that question. Use the current framework for content and the matching released guidelines when checking a particular practice response.
For your next practice task, compare two series with the same radius but different coefficients. Explain every endpoint bracket before checking an answer key. If you want help reviewing that reasoning, see the AP Calculus BC tuition page; a study sequence or tutoring session does not guarantee an exam result.