AP Calculus BC Quiz: Harmonic Series And P Series
20 questions · exam conditions
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Harmonic Series And P SeriesQuestion 1 of 20

A lab totals measurement drift ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p}; for which pp does the series converge?

Converges for p≥1p \geq 1
Converges for p>0p>0
Converges for p>1p>1
Converges for p<1p<1
Converges for p≤1p \leq 1
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AP Calculus BC Quiz

AP Calculus BC Quiz: Harmonic Series And P Series

Practice Harmonic Series And P Series in AP Calculus BC with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Harmonic Series And P Series, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A lab totals measurement drift ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p}; for which pp does the series converge?

  1. Converges for p≥1p \geq 1
  2. Converges for p>0p>0
  3. Converges for p>1p>1 (correct answer)
  4. Converges for p<1p<1
  5. Converges for p≤1p \leq 1
Explanation: This question tests p-series convergence to determine when lab measurement drift converges. The p-series ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} converges if and only if p>1p > 1. For values p≤1p \leq 1, the series diverges because the individual terms don't decrease fast enough to produce convergence. The harmonic series (when p=1p = 1) represents the critical dividing case and diverges. Choice A (p≥1p \geq 1) is tempting because it includes the boundary, but p=1p = 1 results in divergence. When analyzing p-series, always confirm the exponent strictly exceeds 1 for convergence.

Question 2

A machine learning update uses ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p}; for which pp is the total update finite?

  1. Converges for p<1p<1
  2. Converges for p>1p>1 (correct answer)
  3. Converges for p≤1p\le 1
  4. Converges for p>0p>0
  5. Converges for p≥1p\ge 1
Explanation: This question tests p-series convergence to determine when machine learning update totals remain finite. The p-series ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} converges if and only if p>1p > 1. When p≤1p \leq 1, the individual terms decrease too slowly for the infinite sum to converge to a finite value. The boundary case p=1p = 1 produces the harmonic series, which diverges. Choice D (p>0p > 0) might seem plausible since it ensures all terms are positive, but this incorrectly includes divergent cases. For p-series problems, remember that the exponent must strictly exceed 1 for convergence.

Question 3

A telescope tracking error is ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p}; which choice correctly states when it converges?

  1. Converges for p≥1p\ge 1
  2. Converges for p>0p>0
  3. Converges for p<1p<1
  4. Converges for p>1p>1 (correct answer)
  5. Converges for p≤1p\le 1
Explanation: This question tests p-series convergence to determine when telescope tracking error converges. The p-series ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} converges if and only if p>1p > 1. For values p≤1p \leq 1, the series diverges because the individual terms don't decrease fast enough to ensure convergence. The harmonic series (when p=1p = 1) represents the critical dividing case and diverges. Choice A (p≥1p \geq 1) is incorrect because it includes the boundary case p=1p = 1, which results in divergence. When working with p-series, always verify that the exponent is strictly greater than 1 for convergence.

Question 4

A student defines S(p)=∑n=1∞1npS(p)=\sum_{n=1}^{\infty} \frac{1}{n^p}; for which pp does S(p)S(p) converge?

  1. Converges for p≤1p\le 1
  2. Converges for p≥1p\ge 1
  3. Converges for p>0p>0
  4. Converges for p<1p<1
  5. Converges for p>1p>1 (correct answer)
Explanation: This question tests p-series convergence to determine when the student-defined function S(p)S(p) converges. The p-series ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} converges if and only if p>1p > 1. For values p≤1p \leq 1, the series diverges because the individual terms don't decrease rapidly enough to ensure convergence. The harmonic series (when p=1p = 1) serves as the critical case and diverges. Choice B (p≥1p \geq 1) is incorrect because it includes the boundary case p=1p = 1, which results in divergence. When working with p-series, always verify the exponent is strictly greater than 1 for convergence.

Question 5

A sound wave model adds harmonics ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p}; which pp makes the total amplitude finite?

  1. Converges for p>1p>1 (correct answer)
  2. Converges for p≥1p\ge 1
  3. Converges for p>0p>0
  4. Converges for p≤1p\le 1
  5. Converges for p<1p<1
Explanation: This question applies p-series convergence testing to determine when sound wave harmonic totals remain finite. The p-series ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} converges if and only if p>1p > 1. When p≤1p \leq 1, the terms decrease too slowly for the infinite series to converge. At the boundary p=1p = 1, we get the harmonic series which diverges. Choice B (p≥1p \geq 1) might seem reasonable since it includes larger values, but the boundary case p=1p = 1 produces divergence. For p-series convergence problems, remember that the exponent must strictly exceed 1.

Question 6

A student considers ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} as a p-series; which condition on pp guarantees convergence?

  1. Converges for p>0p>0
  2. Converges for p≥1p\ge 1
  3. Converges for p>1p>1 (correct answer)
  4. Converges for p≤1p\le 1
  5. Converges for p<1p<1
Explanation: This question requires p-series convergence analysis to determine when the classic p-series converges. The p-series ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} converges if and only if p>1p > 1. For p≤1p \leq 1, the series diverges because the terms 1np\frac{1}{n^p} don't decrease rapidly enough to produce a finite sum. The critical case p=1p = 1 yields the divergent harmonic series. Choice A (p>0p > 0) is tempting because it includes all positive values, but this incorrectly includes many divergent scenarios. When analyzing p-series, always confirm that the exponent strictly exceeds 1 for convergence.

Question 7

A data logger stores values totaling ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p}; which statement gives the correct convergence rule?

  1. Converges for p>0p>0
  2. Converges for p≤1p\le 1
  3. Converges for p>1p>1 (correct answer)
  4. Converges for p<1p<1
  5. Converges for p≥1p\ge 1
Explanation: This question applies p-series convergence testing to determine when data logger value totals converge. The p-series ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} converges if and only if p>1p > 1. For p≤1p \leq 1, the series diverges because the terms don't decrease rapidly enough to produce a finite sum. At p=1p = 1, we obtain the harmonic series which is known to diverge. Choice A (p>0p > 0) is incorrect because it includes many divergent cases like p=0.5p = 0.5 or p=1p = 1. When working with p-series, the fundamental rule is that the exponent must strictly exceed 1 for convergence.

Question 8

A heat-loss estimate totals ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p}; which statement correctly gives convergence in terms of pp?

  1. Converges for p<1p<1
  2. Converges for p>0p>0
  3. Converges for p≤1p\le 1
  4. Converges for p>1p>1 (correct answer)
  5. Converges for p≥1p\ge 1
Explanation: This question requires p-series convergence analysis to determine when heat-loss estimate totals converge. The p-series ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} converges if and only if p>1p > 1. For p≤1p \leq 1, the series diverges because the terms 1np\frac{1}{n^p} don't decrease fast enough to produce a finite sum. The critical case p=1p = 1 yields the divergent harmonic series. Choice B (p>0p > 0) is incorrect because it includes many divergent scenarios such as p=0.7p = 0.7. When working with p-series, the fundamental rule is that the exponent must strictly exceed 1 for convergence.

Question 9

A drone corrects altitude by ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} meters; which pp keeps total correction finite?

  1. Converges for p≤1p\le 1
  2. Converges for p>0p>0
  3. Converges for p<1p<1
  4. Converges for p>1p>1 (correct answer)
  5. Converges for p≥1p\ge 1
Explanation: This question applies p-series convergence testing to determine when drone altitude corrections remain finite. The p-series ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} converges if and only if p>1p > 1. When p≤1p \leq 1, the terms decrease too slowly for the infinite sum to converge. At the boundary p=1p = 1, we get the harmonic series which diverges. Choice B (p>0p > 0) might appear correct since it ensures positive terms, but this incorrectly includes divergent scenarios like p=0.7p = 0.7. For p-series convergence problems, the key requirement is that the exponent must strictly exceed 1.

Question 10

An algorithm adds step sizes ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p}; which choice correctly states when the sum converges?

  1. Converges for p<1p<1
  2. Converges for p>1p>1 (correct answer)
  3. Converges for p≤1p\le 1
  4. Converges for p>0p>0
  5. Converges for p≥1p\ge 1
Explanation: This question applies p-series convergence testing to determine when algorithm step sizes converge. The p-series ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} converges if and only if p>1p > 1. When p≤1p \leq 1, the terms decrease too slowly for the infinite series to have a finite limit. At the boundary p=1p = 1, we get the harmonic series which diverges. Choice D (p>0p > 0) might seem reasonable since it ensures positive terms, but this incorrectly includes divergent cases like p=0.8p = 0.8. For p-series convergence problems, remember the key requirement: the exponent must exceed 1.

Question 11

A structural damping sum is ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p}; which condition on pp guarantees convergence?

  1. Converges for p>0p>0
  2. Converges for p≤1p\le 1
  3. Converges for p<1p<1
  4. Converges for p≥1p\ge 1
  5. Converges for p>1p>1 (correct answer)
Explanation: This question applies p-series convergence testing to determine when structural damping sums converge. The p-series ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} converges if and only if p>1p > 1. When p≤1p \leq 1, the terms decrease too slowly for the infinite series to have a finite limit. At the boundary p=1p = 1, we get the harmonic series which diverges. Choice A (p>0p > 0) might seem plausible since it ensures positive terms, but this incorrectly includes divergent cases like p=0.7p = 0.7. For p-series convergence problems, remember that the exponent must strictly exceed 1.

Question 12

A cooling model sums ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p}; which statement correctly describes when the series converges?

  1. Converges for p>1p>1 (correct answer)
  2. Converges for p≤1p\le 1
  3. Converges for p<1p<1
  4. Converges for p>0p>0
  5. Converges for p≥1p\ge 1
Explanation: This question requires p-series convergence analysis to determine when the cooling model series converges. The p-series ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} converges if and only if p>1p > 1. When p≤1p \leq 1, the terms decrease too slowly for the infinite sum to converge to a finite value. The harmonic series (when p=1p = 1) serves as the critical dividing case and diverges. Choice E (p≥1p \geq 1) might appear correct since it includes larger values, but the boundary case p=1p = 1 produces divergence. For any p-series problem, verify that the exponent strictly exceeds 1 for convergence.

Question 13

A model for friction loss uses ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p}; for which pp does the loss converge?

  1. Converges for p<1p<1
  2. Converges for p>1p>1 (correct answer)
  3. Converges for p>0p>0
  4. Converges for p≥1p\ge 1
  5. Converges for p≤1p\le 1
Explanation: This question requires p-series convergence analysis to determine when friction loss models converge. The p-series ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} converges if and only if p>1p > 1. When p≤1p \leq 1, the terms 1np\frac{1}{n^p} decrease too slowly for the infinite sum to have a finite limit. The critical boundary at p=1p = 1 gives the divergent harmonic series. Choice C (p>0p > 0) might seem plausible since it ensures positive terms, but this incorrectly includes divergent scenarios such as p=0.8p = 0.8. For p-series convergence problems, remember the key criterion: the exponent must exceed 1.

Question 14

A hiker's elevation gains sum ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} meters; which pp keeps total gain finite?

  1. Converges for p>0p>0
  2. Converges for p<1p<1
  3. Converges for p≥1p\ge 1
  4. Converges for p≤1p\le 1
  5. Converges for p>1p>1 (correct answer)
Explanation: This question applies p-series convergence testing to determine when hiker elevation gain totals remain finite. The p-series ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} converges if and only if p>1p > 1. For p≤1p \leq 1, the terms don't decrease rapidly enough for the infinite series to converge. At p=1p = 1, we obtain the harmonic series which is known to diverge. Choice A (p>0p > 0) might appear plausible since it ensures positive terms, but this incorrectly includes many divergent cases like p=0.8p = 0.8. When analyzing p-series convergence, always confirm the exponent strictly exceeds 1.

Question 15

A population model adds effects as ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p}; which choice gives the correct convergence condition?

  1. Converges for p≤1p\le 1
  2. Converges for p>1p>1 (correct answer)
  3. Converges for p≥1p\ge 1
  4. Converges for p>0p>0
  5. Converges for p<1p<1
Explanation: This question tests p-series convergence to determine when population model effects converge. The p-series ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} converges if and only if p>1p > 1. When p≤1p \leq 1, the individual terms 1np\frac{1}{n^p} decrease too slowly for the infinite series to converge. The boundary case p=1p = 1 produces the harmonic series, which is known to diverge. Choice D (p>0p > 0) might appear reasonable since it ensures all terms are positive, but this incorrectly includes many divergent cases. For p-series problems, always check that the exponent is strictly greater than 1 for convergence.

Question 16

A student tests ∑n=1∞1n9/8\sum_{n=1}^{\infty} \frac{1}{n^{9/8}}; which statement about convergence is correct?

  1. Diverges because p<1p<1.
  2. Converges because p>1p>1. (correct answer)
  3. Converges because p=1p=1.
  4. Diverges because p>1p>1.
  5. Converges because p≤1p\le 1.
Explanation: The student is testing the p-series ∑n=1∞1n9/8\sum_{n=1}^{\infty} \frac{1}{n^{9/8}} for convergence. Here, p=9/8=1.125p = 9/8 = 1.125, which is greater than 1. The p-series test tells us that ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} converges when p>1p > 1 and diverges when p≤1p \leq 1. Since 9/8>19/8 > 1, this series converges. Choice A incorrectly claims divergence because p<1p < 1, but 9/8=1.125>19/8 = 1.125 > 1, demonstrating a misunderstanding of fraction comparison. For p-series with fractional exponents, check if the numerator exceeds the denominator—if yes, then p>1p > 1 and the series converges.

Question 17

In a model, decide whether ∑n=1∞1n4/3\sum_{n=1}^{\infty} \frac{1}{n^{4/3}} converges or diverges.

  1. Converges because p=1p=1.
  2. Diverges because p>1p>1.
  3. Diverges because p<1p<1.
  4. Converges because p>1p>1. (correct answer)
  5. Converges because p≤1p\le 1.
Explanation: This model requires analyzing the p-series ∑n=1∞1n4/3\sum_{n=1}^{\infty} \frac{1}{n^{4/3}} for convergence. The exponent p=4/3≈1.333p = 4/3 \approx 1.333 is greater than 1. According to the p-series test, a series of the form ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} converges if and only if p>1p > 1. Since 4/3>14/3 > 1, this series converges. Choice C incorrectly states the series diverges because p<1p < 1, but 4/3=1.333...>14/3 = 1.333... > 1, showing a fundamental error in comparing fractions. When testing p-series, convert fractions to decimals if needed to clearly see whether p>1p > 1 for convergence.

Question 18

A sensor records ∑n=1∞1n7/6\sum_{n=1}^{\infty} \frac{1}{n^{7/6}}; does the series converge or diverge?

  1. Diverges because p=1p=1.
  2. Converges because p>1p>1. (correct answer)
  3. Diverges because p>1p>1.
  4. Converges because p≤1p\le 1.
  5. Diverges because p<1p<1.
Explanation: This sensor recording problem involves testing the p-series ∑n=1∞1n7/6\sum_{n=1}^{\infty} \frac{1}{n^{7/6}}. Here, p=7/6≈1.167p = 7/6 \approx 1.167, which is greater than 1. The p-series convergence test states that ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} converges when p>1p > 1 and diverges when p≤1p \leq 1. Since 7/6>17/6 > 1, this series converges. Choice A incorrectly claims divergence because p=1p = 1, but 7/6≠17/6 \neq 1, revealing confusion about fraction evaluation. To quickly check p-series convergence, compare the exponent to 1: if the exponent exceeds 1, the series converges.

Question 19

A telescope adds blur corrections ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p}; for which pp is the total correction finite?

  1. Converges for p>0p>0
  2. Converges for p<1p<1
  3. Converges for p≥1p\ge 1
  4. Converges for p>1p>1 (correct answer)
  5. Converges for p≤1p\le 1
Explanation: This question tests p-series convergence to determine when telescope blur corrections remain finite. The p-series ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} converges if and only if p>1p > 1. For p≤1p \leq 1, the series diverges because the individual terms don't decrease rapidly enough to ensure convergence. The boundary case p=1p = 1 gives the harmonic series, which is a classic example of divergence. Choice A (p>0p > 0) is incorrect because it includes many divergent cases like p=0.5p = 0.5 or p=1p = 1. When working with p-series, always confirm the exponent is strictly greater than 1 for convergence.

Question 20

A savings plan deposits ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} dollars; for which pp does the total deposited converge?

  1. Converges for p>0p>0
  2. Converges for p>1p>1 (correct answer)
  3. Converges for p≥0p\ge 0
  4. Converges for p<1p<1
  5. Converges for p≤1p\le 1
Explanation: This question requires applying p-series convergence testing to determine when the savings deposits converge. The p-series ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} converges if and only if p>1p > 1. When p≤1p \leq 1, the terms 1np\frac{1}{n^p} decrease too slowly for the infinite sum to have a finite value. For p=1p = 1, we get the harmonic series which famously diverges. Choice A (p>0p > 0) is tempting because it includes all positive values, but this incorrectly includes cases like p=12p = \frac{1}{2} where the series diverges. To determine p-series convergence, always check whether the exponent exceeds 1.