Free Printable WorksheetMathCalculus IIMath CurriculumAP Calculus BC Unit 10; CCSS.HSA.SSE, HSF.BF; CCSS.MP2, MP5, MP6, MP7, MP8; Integrated Calculus II Standards CAL2-PS; TEKS 19 TAC section 111.55

CAL2.138 Endpoint Testing for Power Series Worksheet

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Worksheet Overview & Skills Covered

Core academic competencies and curriculum alignment

This Calculus II Math worksheet provides structured practice on CAL2.138 Endpoint Testing for Power Series. Students are guided through targeted exercises that build foundational understanding, computational accuracy, and critical thinking skills.

CAL2.138 Endpoint Testing for Power Series - Choose the convergence conclusion at each endpoint of an interval of convergence. Standards tags: AP Calculus BC Unit 10; CCSS.HSA.SSE, HSF.BF; CCSS.MP2, MP5, MP6, MP7, MP8; Integrated Calculus II Standards CAL2-PS; TEKS 19 TAC section 111.55.

Whether you are an educator preparing classroom handouts, a tutor reinforcing weekly topics, or a parent supporting homeschool learning, this printable resource gives students the deliberate practice required to achieve subject fluency and master state standards.

Grade

Calculus II

Subject

Math

Domain

Math Curriculum

Formats

PDF & DOCX

Sample Practice Problems

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Below are sample questions representative of the exercises included on this worksheet and related CAL2.138 Endpoint Testing for Power Series skill assessments:

Problem 1Difficulty: Mastery
For the power series S(x) = Σ_{n=2}^∞ [1/(n(ln(n))^(3/2))]((x - 6)/2)n, the open interval from the ratio/root test is 4 < x < 8. After testing both endpoints, which interval of convergence is correct?
A(4, 8)
B[4, 8)
C(4, 8]
D[4, 8]
Solution & Explanation:At x = 4, the factor ((x - 6)/2)n becomes (-1)n, so the endpoint series alternates. The endpoint series is Σ_{n=2}^∞ (-1)n·1/(n(ln(n))^(3/2)), which converges absolutely by the integral test because the logarithmic exponent 3/2 is greater than 1. At x = 8, the factor ((x - 6)/2)n becomes 1, so the endpoint series is positive. The endpoint series is Σ_{n=2}^∞ 1/(n(ln(n))^(3/2)), which converges absolutely by the integral test because the logarithmic exponent 3/2 is greater than 1. Therefore the interval includes exactly the endpoints that converge: [4, 8]. A logarithmic exponent greater than 1 is enough for positive endpoint convergence.
Problem 2Difficulty: Mastery
A student tests S(x) = Σ_{n=1}^∞ [n]((x + 4)/8)n and finds the interior interval -12 < x < 4. The student says, 'The endpoint behavior should match on both sides because both endpoints are 8 units from the center.' Which correction is accurate?
Ax = -12: converges absolutely; x = 4: diverges
Bx = -12: diverges; x = 4: converges absolutely
Cx = -12: diverges; x = 4: diverges
Dx = -12: converges absolutely; x = 4: converges conditionally
Solution & Explanation:At x = -12, the factor ((x + 4)/8)n becomes (-1)n, so the endpoint series alternates. The endpoint series is Σ_{n=1}^∞ (-1)n·n, which diverges because the terms do not approach 0 even with alternating signs. At x = 4, the factor ((x + 4)/8)n becomes 1, so the endpoint series is positive. The endpoint series is Σ_{n=1}^∞ n, which diverges because the terms do not approach 0. The ratio or root test only gives the open interval; endpoints must be tested separately. The nth-term test must be checked before using any alternating-series reasoning.
Problem 3Difficulty: Mastery
For S(x) = Σ_{n=1}^∞ [(-1)n·n2/(n2 + 1)]((x - 4)/7)n, endpoint substitution gives boundary series at x = -3 and x = 11. Which option uses the correct endpoint tests and conclusions?
AAt both endpoints, the ratio test gives a limit less than 1, so both endpoints converge automatically.
BAt x = -3, diverges because n2/(n2 + 1) approaches 1, so terms do not approach 0; at x = 11, diverges because the term size approaches 1 even with alternating signs.
CAt x = -3, ignore the sign and use only the positive-series test; at x = 11, do the same.
DBoth endpoints diverge because the ratio test is inconclusive when its limit equals 1.
Solution & Explanation:At x = -3, the factor ((x - 4)/7)n becomes (-1)n, and the coefficient's (-1)n cancels it to give positive terms. The endpoint series is Σ_{n=1}^∞ n2/(n2 + 1), which diverges because n2/(n2 + 1) approaches 1, so terms do not approach 0. At x = 11, the factor ((x - 4)/7)n becomes 1, so the coefficient's (-1)n remains. The endpoint series is Σ_{n=1}^∞ (-1)n·n2/(n2 + 1), which diverges because the term size approaches 1 even with alternating signs. The ratio or root test only gives the open interval; endpoints must be tested separately. A rational expression with equal highest powers does not produce terms tending to 0.

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QWhat grade level and standards does this worksheet align with?

This worksheet is designed specifically for Calculus II Math students. It aligns with educational standard AP Calculus BC Unit 10; CCSS.HSA.SSE, HSF.BF; CCSS.MP2, MP5, MP6, MP7, MP8; Integrated Calculus II Standards CAL2-PS; TEKS 19 TAC section 111.55 (Choose the convergence conclusion at each endpoint of an interval of convergence. Standards tags: AP Calculus BC Unit 10; CCSS.HSA.SSE, HSF.BF; CCSS.MP2, MP5, MP6, MP7, MP8; Integrated Calculus II Standards CAL2-PS; TEKS 19 TAC section 111.55.).

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