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  3. AP Calculus BC

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AP Calculus BC Flashcards: Harmonic Series And P Series

Study Harmonic Series And P Series in AP Calculus BC with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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What this deck covers

This deck focuses on Harmonic Series And P Series, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus BC.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

AP Calculus BC Flashcards: Harmonic Series And P Series

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QUESTION

What condition makes a p-series diverge?

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ANSWER

Diverges if p≤1p \leq 1p≤1. When p≤1p \leq 1p≤1, terms don't decrease fast enough.

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All flashcards

Flashcard 1: What condition makes a p-series diverge?

Answer: Diverges if p≤1p \leq 1p≤1. When p≤1p \leq 1p≤1, terms don't decrease fast enough.

Flashcard 2: Does the series ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{n}∑n=1∞​n1​ converge or diverge?

Answer: Diverges. The harmonic series is the classic divergent series.

Flashcard 3: State the formula for a p-series.

Answer: ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p}∑n=1∞​np1​. General form where ppp determines convergence behavior.

Flashcard 4: Does ∑n=1∞1n0.7\sum_{n=1}^{\infty} \frac{1}{n^{0.7}}∑n=1∞​n0.71​ converge or diverge?

Answer: Diverges. Since p=0.7<1p = 0.7 < 1p=0.7<1, this p-series diverges.

Flashcard 5: Is ∑n=1∞1n0.5\sum_{n=1}^{\infty} \frac{1}{n^{0.5}}∑n=1∞​n0.51​ convergent or divergent?

Answer: Divergent. Since p=0.5≤1p = 0.5 \leq 1p=0.5≤1, this p-series diverges.

Flashcard 6: Identify the convergence of ∑n=1∞1n2\sum_{n=1}^{\infty} \frac{1}{n^2}∑n=1∞​n21​.

Answer: Converges. Since p=2>1p = 2 > 1p=2>1, this p-series converges.

Flashcard 7: What is the nature of ∑n=1∞1n2.1\sum_{n=1}^{\infty} \frac{1}{n^{2.1}}∑n=1∞​n2.11​?

Answer: Convergent. Since p=2.1>1p = 2.1 > 1p=2.1>1, this p-series converges.

Flashcard 8: What is the definition of a harmonic series?

Answer: The series ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{n}∑n=1∞​n1​. The classic divergent series with terms 1n\frac{1}{n}n1​.

Flashcard 9: Identify if the series ∑n=1∞1n1.5\sum_{n=1}^{\infty} \frac{1}{n^{1.5}}∑n=1∞​n1.51​ converges.

Answer: Converges. Since p=1.5>1p = 1.5 > 1p=1.5>1, this p-series converges.

Flashcard 10: What condition makes a p-series converge?

Answer: Converges if p>1p > 1p>1. When p>1p > 1p>1, the terms decrease fast enough for convergence.

Flashcard 11: Does the series ∑n=1∞1n3.5\sum_{n=1}^{\infty} \frac{1}{n^{3.5}}∑n=1∞​n3.51​ converge?

Answer: Converges. Since p=3.5>1p = 3.5 > 1p=3.5>1, this p-series converges.

Flashcard 12: Identify the type of series: ∑n=1∞1n0.9\sum_{n=1}^{\infty} \frac{1}{n^{0.9}}∑n=1∞​n0.91​.

Answer: p-series. Has form ∑1np\sum \frac{1}{n^p}∑np1​ with p=0.9p = 0.9p=0.9.

Flashcard 13: State the convergence of ∑n=1∞1n0.8\sum_{n=1}^{\infty} \frac{1}{n^{0.8}}∑n=1∞​n0.81​.

Answer: Diverges. Since p=0.8<1p = 0.8 < 1p=0.8<1, this p-series diverges.

Flashcard 14: Determine the convergence of ∑n=1∞1n4\sum_{n=1}^{\infty} \frac{1}{n^{4}}∑n=1∞​n41​.

Answer: Converges. Since p=4>1p = 4 > 1p=4>1, this p-series converges.

Flashcard 15: What is the second term of a harmonic series?

Answer: 12\frac{1}{2}21​. The second term in the harmonic series is 12\frac{1}{2}21​.

Flashcard 16: Identify if ∑n=1∞1n1/2\sum_{n=1}^{\infty} \frac{1}{n^{1/2}}∑n=1∞​n1/21​ converges or diverges.

Answer: Diverges. Since p=0.5<1p = 0.5 < 1p=0.5<1, this p-series diverges.

Flashcard 17: Identify if the series ∑n=1∞1n0.6\sum_{n=1}^{\infty} \frac{1}{n^{0.6}}∑n=1∞​n0.61​ is convergent.

Answer: Diverges. Since p=0.6<1p = 0.6 < 1p=0.6<1, this p-series diverges.

Flashcard 18: Identify if ∑n=1∞1n3/2\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}∑n=1∞​n3/21​ converges or diverges.

Answer: Converges. Since p=1.5>1p = 1.5 > 1p=1.5>1, this p-series converges.

Flashcard 19: Identify the series that converges: ∑n=1∞1n3\sum_{n=1}^{\infty} \frac{1}{n^3}∑n=1∞​n31​ or ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{n}∑n=1∞​n1​.

Answer: ∑n=1∞1n3\sum_{n=1}^{\infty} \frac{1}{n^3}∑n=1∞​n31​. Since p=3>1p = 3 > 1p=3>1, the first series converges.

Flashcard 20: Which series is divergent: ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{n}∑n=1∞​n1​ or ∑n=1∞1n2\sum_{n=1}^{\infty} \frac{1}{n^2}∑n=1∞​n21​?

Answer: ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{n}∑n=1∞​n1​. The harmonic series diverges while ∑1n2\sum \frac{1}{n^2}∑n21​ converges.

Flashcard 21: State whether ∑n=1∞1n2.5\sum_{n=1}^{\infty} \frac{1}{n^{2.5}}∑n=1∞​n2.51​ converges.

Answer: Converges. Since p=2.5>1p = 2.5 > 1p=2.5>1, this p-series converges.

Flashcard 22: For which ppp is the series ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p}∑n=1∞​np1​ divergent?

Answer: p≤1p \leq 1p≤1. P-series diverge when ppp is at most 1.

Flashcard 23: What is the fifth term of a harmonic series?

Answer: 15\frac{1}{5}51​. The fifth term in the harmonic series is 15\frac{1}{5}51​.

Flashcard 24: What is the fourth term of a harmonic series?

Answer: 14\frac{1}{4}41​. The fourth term in the harmonic series is 14\frac{1}{4}41​.

Flashcard 25: What is the condition for the convergence of a harmonic series?

Answer: Always diverges. The harmonic series is famously always divergent.

Flashcard 26: What is the third term of a harmonic series?

Answer: 13\frac{1}{3}31​. The third term in the harmonic series is 13\frac{1}{3}31​.

Flashcard 27: What is the first term of a harmonic series?

Answer:

  1. The harmonic series starts with 11=1\frac{1}{1} = 111​=1.

Flashcard 28: What type of series is ∑n=1∞1n0.3\sum_{n=1}^{\infty} \frac{1}{n^{0.3}}∑n=1∞​n0.31​?

Answer: p-series. Has form ∑1np\sum \frac{1}{n^p}∑np1​ with p=0.3p = 0.3p=0.3.

Flashcard 29: What type of series is ∑n=1∞1n1.5\sum_{n=1}^{\infty} \frac{1}{n^{1.5}}∑n=1∞​n1.51​?

Answer: p-series. Has the form ∑1np\sum \frac{1}{n^p}∑np1​ with p=1.5p = 1.5p=1.5.

Flashcard 30: What is the definition of a harmonic series?

Answer: The series ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{n}∑n=1∞​n1​. The classic divergent series with terms 1n\frac{1}{n}n1​.