--- title: "Power Series and the Radius of Convergence" description: "Read a power series as a function defined where it converges, find its radius with the ratio test, and check the endpoints separately." canonical: https://lightmysky.com/learn/mathematics/power-series-and-the-radius-of-convergence-mt_RnYb0JLKbD source: https://lightmysky.com/learn/mathematics/power-series-and-the-radius-of-convergence-mt_RnYb0JLKbD.md retrieved: 2026-11-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # Power Series and the Radius of Convergence Read a power series as a function defined where it converges, find its radius with the ratio test, and check the endpoints separately. Subject: Mathematics · Area: Calculus & Analysis · Ages 19 to 20 Page: https://lightmysky.com/learn/mathematics/power-series-and-the-radius-of-convergence-mt_RnYb0JLKbD ## Ready when they can - Find the radius of convergence of a given power series - Test both endpoints and state the full interval - Differentiate or integrate a power series term by term inside its radius ## Lesson: How far a power series reaches A power series centered at c is a sum of terms a_n(x-c)^n, for n = 0, 1, 2, ... The coefficients a_n are fixed numbers. The n = 0 term is a_0, even at x = c. For finite R > 0, it converges absolutely when |x-c| < R and diverges when |x-c| > R. The boundaries need separate tests. R = 0 means only the center converges. R = infinity means the series converges for every real x. Fix x and write u_n for the nth term. When consecutive ratios are eventually defined, find L = lim |u_(n+1)/u_n| as n grows. L < 1 proves absolute convergence. L > 1, including infinity, proves divergence. L = 1 gives no conclusion. Check the center directly, since its terms may make the ratio undefined. The ratio test does not settle every power series. For terms (b(x-c))^n/n^p, with n >= 1, b nonzero and fixed p >= 0, the absolute ratio is |b(x-c)|(n/(n+1))^p away from c. Its limit is |b(x-c)|, giving R = 1/|b|. For p = 0 this is geometric: convergence holds exactly when |b(x-c)| < 1. At either boundary its terms have magnitude 1, so they do not tend to zero. **Example.** You have tested alternating series before. For sum (x+1)^n/(n*3^n), n >= 1, the ratio limit is |x+1|/3. Thus c = -1, R = 3 and the interior is -4 < x < 2. At -4 the terms are (-1)^n/n; the series converges by the alternating test. At 2 the terms are 1/n; the series diverges. The interval is [-4, 2). A square bracket includes a boundary; a parenthesis excludes it. *(drawing: The center is -1. Each boundary is distance 3 from it. The interval is [-4, 2).)* Here n! means the product 1*2*...*n. For terms n!(x-c)^n, n >= 1, the ratio is (n+1)|x-c| away from c. It tends to infinity, so only the center converges and R = 0. For terms (x-c)^n/n!, the ratio is |x-c|/(n+1). It tends to 0 for every fixed x away from c. The center also converges, so R = infinity. Suppose the nonzero coefficients satisfy |a_(n+1)/a_n| -> A, where A is finite and positive. For sum a_n(x-c)^n, the term-ratio limit away from c is A|x-c|. Solve A|x-c| < 1 to get |x-c| < 1/A. Thus R = 1/A. Measure from the center to either boundary, not across the whole interval. Differentiation finds a function's rate of change; f'(x) names its derivative. For a power series with R > 0, differentiate each term inside |x-c| < R: a_n(x-c)^n becomes n*a_n(x-c)^(n-1), for n >= 1. The constant term becomes 0. The differentiated series has the same radius. Check its endpoints again: terms that do not tend to zero prove divergence. An antiderivative F of f satisfies F' = f. For continuous f, its definite integral from c to x is F(x)-F(c), which is 0 at x = c. Inside the radius, integrate a_n(x-c)^n to get a_n(x-c)^(n+1)/(n+1). Add C for an antiderivative; integrating from c to x gives C = 0 in this form. The new series keeps its radius; test its endpoints again. For a coefficient of power k, choose the original power k-1. **Example.** The geometric series with terms (x-1)^n/4^n, n >= 0, converges exactly when |x-1| < 4, so R = 4. Integrating from 1 to x gives terms (x-1)^(n+1)/((n+1)*4^n). Its radius remains 4. For instance, its quadratic term comes from n = 1 and is (x-1)^2/8. **Recap.** Find the radius from a convergence test, check finite endpoints separately, and retain the radius after term-by-term differentiation or integration. *This lesson, quiz and flashcards were written for this learning dataset.* ## Practice 14 questions on this page, each with its working shown. ## Needs first - [Alternating Series and Absolute Convergence](https://lightmysky.com/learn/mathematics/alternating-series-and-absolute-convergence-mt_sk6F4n1USH) ## Optional connections - [Infinite Series and the Geometric Series](https://lightmysky.com/learn/mathematics/infinite-series-and-the-geometric-series-mt_EPE-ERFFQt) ## Opens up - [Pointwise and Uniform Convergence](https://lightmysky.com/learn/mathematics/pointwise-and-uniform-convergence-mt_iAGk1ML5MA) - [Power Series and Analyticity in the Complex Plane](https://lightmysky.com/learn/mathematics/power-series-and-analyticity-in-the-complex-plane-mt_Mwxapy_pU8) - [Taylor and Maclaurin Series](https://lightmysky.com/learn/mathematics/taylor-and-maclaurin-series-mt_SgI9Pn9RIO) - [Generating Functions](https://lightmysky.com/learn/mathematics/generating-functions-mt_WAX3O-HjRk)