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How do you find the radius and interval of convergence of a power series?

Topic 10.13 Radius and Interval of Convergence of Power Series: find the radius and interval of convergence of a power series using the ratio test and checking the endpoints (BC).

A focused answer to AP Calculus BC Topic 10.13, finding the radius and interval of convergence of a power series with the ratio test and separately testing the endpoints, with worked examples.

Generated by Claude Opus 4.810 min answer

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  1. What this topic is asking
  2. The method
  3. Why the ratio test gives the radius
  4. A worked radius and interval
  5. The endpoints are where the work is
  6. The three possible radii

What this topic is asking

The College Board (Topic 10.13, BC only) studies power series βˆ‘cn(xβˆ’a)n\sum c_n(x - a)^n, which are "infinite polynomials" in xx. A power series converges for some values of xx and diverges for others; this topic finds the interval of xx-values where it converges, using the ratio test and a careful check of the two endpoints.

The method

Why the ratio test gives the radius

The ratio test is the natural tool because a power series has the nn-th-power structure the ratio test handles best. Forming ∣cn+1(xβˆ’a)n+1cn(xβˆ’a)n∣\left|\frac{c_{n+1}(x-a)^{n+1}}{c_n(x-a)^n}\right|, the powers combine to a single factor ∣xβˆ’a∣|x - a|, leaving ∣xβˆ’aβˆ£β‹…lim⁑∣cn+1cn∣|x - a|\cdot\lim\left|\frac{c_{n+1}}{c_n}\right|. Setting this <1< 1 for absolute convergence isolates ∣xβˆ’a∣|x - a| below a threshold RR, the radius. So the ratio test does not just decide one series; for a power series it produces an inequality in xx that defines the convergence set. The endpoints, where the ratio-test limit equals exactly 11, are precisely the inconclusive case, which is why they always need separate handling.

A worked radius and interval

The endpoints are where the work is

Finding the radius is usually quick; the endpoints are where most of the marks and most of the errors live. At each endpoint the ratio-test limit is exactly 11, so the ratio test says nothing, and you must substitute the endpoint xx-value back into the series and apply a different test. Common outcomes: a pp-series, the harmonic series (diverges), an alternating series (often converges by the alternating series test), or a geometric series. Each endpoint can behave differently, as the worked example shows, so test them independently. The final interval includes an endpoint with a square bracket only if the series converges there. Skipping endpoint analysis, or assuming both endpoints behave the same, is the classic loss of credit.

The three possible radii

Every power series falls into exactly one of three cases for RR. If the ratio-test limit is 00 for all xx (as for βˆ‘xnn!\sum\frac{x^n}{n!}, since the factorial dominates), the series converges everywhere and R=∞R = \infty; there are no endpoints to test. If the limit is ∞\infty for all xβ‰ ax\neq a (as for βˆ‘n! xn\sum n!\,x^n), the series converges only at the center x=ax = a and R=0R = 0. Otherwise 0<R<∞0 < R < \infty and you get a finite interval requiring endpoint analysis. Recognizing which case you are in from the ratio-test limit tells you immediately whether endpoint work is even needed: only the finite-RR case has endpoints.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2021 (BC, style)1 marksSection I (multiple choice, no calculator). The radius of convergence of βˆ‘n=0∞xnn!\sum_{n=0}^{\infty} \frac{x^n}{n!} is (A) ∞\infty (B) 11 (C) 00 (D) ee
Show worked answer β†’

The correct answer is (A), ∞\infty.

Ratio test: ∣xn+1/(n+1)!xn/n!∣=∣x∣n+1β†’0<1\left|\frac{x^{n+1}/(n+1)!}{x^n/n!}\right| = \frac{|x|}{n+1}\to 0 < 1 for every xx. The series converges for all xx, so the radius of convergence is ∞\infty.

AP 2024 (BC, style)4 marksSection II (free response, no calculator). Find the radius and interval of convergence of βˆ‘n=1∞(xβˆ’2)nn\sum_{n=1}^{\infty} \frac{(x - 2)^n}{n}, including endpoint analysis.
Show worked answer β†’

A 4-point radius-and-interval problem.

(2 points) Ratio test: ∣(xβˆ’2)n+1/(n+1)(xβˆ’2)n/n∣=∣xβˆ’2βˆ£β‹…nn+1β†’βˆ£xβˆ’2∣\left|\frac{(x-2)^{n+1}/(n+1)}{(x-2)^n/n}\right| = |x-2|\cdot\frac{n}{n+1}\to |x - 2|. Converges when ∣xβˆ’2∣<1|x - 2| < 1, so radius R=1R = 1 and the open interval is 1<x<31 < x < 3.
(2 points) At x=3x = 3: βˆ‘1n\sum\frac{1}{n} diverges. At x=1x = 1: βˆ‘(βˆ’1)nn\sum\frac{(-1)^n}{n} converges (alternating). Interval of convergence: [1,3)[1, 3).

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