AP Calculus BC: Radius, Interval and Types of Convergence

To find a power series’ interval of convergence, first determine its radius, then test the endpoints separately. This guide works through both steps and explains how to distinguish absolute convergence, conditional convergence and divergence.

Radius and interval of convergence

A power series has the form ∑ aₙ(x − c)ⁿ, where c is its centre. Its radius of convergence R tells you how far from c it converges absolutely: |x − c| < R. It diverges when |x − c| > R. When 0 < R < ∞, test x = c − R and x = c + R individually to decide whether either endpoint belongs to the interval. If R = 0, only the centre is included; if R = ∞, the series converges for every real x.

Use the ratio test, then check its limits

For a series ∑ uₙ with eventually nonzero terms, calculate L = lim |uₙ₊₁/uₙ| as n → ∞, if the limit exists. If L < 1, the series converges absolutely. If L > 1, including an infinite limit, it diverges. If L = 1, the test is inconclusive: use another test. For power series, keep x fixed while taking the limit and check the centre separately if forming the ratio divides by zero.

Worked example: one endpoint included

Consider ∑ (x − 3)ⁿ/(n·2ⁿ), starting at n = 1. Away from x = 3, the ratio of successive terms in absolute value is [n/(n + 1)]·|x − 3|/2. Its limit is |x − 3|/2, so absolute convergence requires |x − 3| < 2. The radius is 2 and the open interval is (1, 5). At the centre, every term is zero.

• At x = 1, the series is ∑ (−1)ⁿ/n. It converges by the alternating series test because 1/n decreases to zero.
• At x = 5, the series is ∑ 1/n, the divergent harmonic series.

The interval of convergence is therefore [1, 5).

Absolute versus conditional convergence

A series ∑ uₙ converges absolutely if ∑ |uₙ| converges. It converges conditionally if ∑ uₙ converges but ∑ |uₙ| diverges. These are distinct classifications: an absolutely convergent series is not conditionally convergent.

In the first example, convergence is absolute for 1 < x < 5. At x = 1, removing the signs gives the divergent harmonic series, so convergence there is conditional. The alternating series test alone does not establish absolute convergence.

Worked example: neither endpoint included

Consider ∑ n(x + 2)ⁿ/4ⁿ, starting at n = 1. Away from x = −2, the ratio-test limit is |x + 2|/4. The inequality |x + 2| < 4 gives −6 < x < 2. The centre is −2 and the radius is 4.

• At x = −6, the terms are n(−1)ⁿ and do not tend to zero, so the series diverges.
• At x = 2, the terms are n and do not tend to zero, so the series diverges.

The interval is (−6, 2). The centre x = −2 is inside the interval, not an endpoint; there, every term is zero.

Check your answer

State the radius, identify both endpoints and justify their inclusion or exclusion. A term limit of zero alone does not prove convergence. Match your final answer to what the question asks; consult the official scoring guidelines for that particular exam question rather than assuming a fixed number of points for each step.

For help with series and other BC topics, explore our AP tutoring courses.

Curriculum reference checked 14 October 2026: College Board AP Calculus AB and BC Course and Exam Description, Unit 10, Topics 10.8, 10.9 and 10.13. The examples above are worked teaching examples, not past exam questions.