Free Printable WorksheetMathCalculus IIMath CurriculumAP Calculus BC Unit 10; CCSS.HSA.SSE, HSF.IF; CCSS.MP2, MP3, MP6, MP7, MP8; Integrated Calculus II Standards CAL2-SER; TEKS 19 TAC section 111.55

CAL2.117 Harmonic 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 CAL2.117 Harmonic Series. Students are guided through targeted exercises that build foundational understanding, computational accuracy, and critical thinking skills.

CAL2.117 Harmonic Series - Select the correct divergence conclusion and explanation for the harmonic series. Standards: AP Calculus BC Unit 10; CCSS.HSA.SSE, HSF.IF; CCSS.MP2, MP3, MP6, MP7, MP8; Integrated Calculus II Standards CAL2-SER; 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

Preview the types of problems and questions covered in this worksheet

Below are sample questions representative of the exercises included on this worksheet and related CAL2.117 Harmonic Series skill assessments:

Problem 1Difficulty: Mastery

Which answer gives the best final conclusion for a proof that has shown H_N ≥ 1 + (1/2)⌊log_2 N⌋?

ASince log_2 N grows slowly, H_N must converge.

BSince the floor function is sometimes constant, the harmonic series converges.

CSince H_N has a lower bound, it must converge.

DSince ⌊log_2 N⌋ grows without bound, H_N is unbounded and ∑ 1/n diverges.

Solution & Explanation:The lower bound itself grows without bound as N grows. Therefore the harmonic partial sums cannot have a finite limit, which proves divergence.

Problem 2Difficulty: Mastery

Which response correctly analyzes the claim 'Since ∑ 1/n diverges, any series with terms less than 1/n must converge'?

ATrue; this is the direct comparison test.

BFalse only because 1/(n+1) is greater than 1/n.

CTrue only for positive decreasing terms.

DFalse; being below a divergent benchmark is not enough, and even ∑ 1/(n+1) is less than 1/n term-by-term but diverges.

Solution & Explanation:A divergent upper comparison gives no convergence guarantee. The shifted harmonic tail is a direct counterexample: it is smaller than 1/n but still diverges.

Problem 3Difficulty: Mastery

Which answer correctly classifies ∑_{n=1}^∞ (n+4)/(n²+8n+16)?

AIt converges because the denominator is a perfect square.

BIt converges because the numerator cancels all terms.

CIt diverges because the terms approach 1.

DIt diverges because (n+4)/(n+4)² = 1/(n+4), a harmonic tail.

Solution & Explanation:The denominator factors as (n+4)², so the term simplifies to 1/(n+4). That is a shifted harmonic tail, which diverges.

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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.IF; CCSS.MP2, MP3, MP6, MP7, MP8; Integrated Calculus II Standards CAL2-SER; TEKS 19 TAC section 111.55 (Select the correct divergence conclusion and explanation for the harmonic series. Standards: AP Calculus BC Unit 10; CCSS.HSA.SSE, HSF.IF; CCSS.MP2, MP3, MP6, MP7, MP8; Integrated Calculus II Standards CAL2-SER; TEKS 19 TAC section 111.55).