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AP Calculus BC Flashcards: Ratio Test For Convergence

Study Ratio Test For Convergence 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 Ratio Test For Convergence, 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: Ratio Test For Convergence

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QUESTION

What does L=1L = 1L=1 imply about the convergence of the series?

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ANSWER

The test is inconclusive. The boundary case provides no information about convergence.

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Flashcard 1: What does L=1L = 1L=1 imply about the convergence of the series?

Answer: The test is inconclusive. The boundary case provides no information about convergence.

Flashcard 2: Determine the Ratio Test result for an=n3e−na_n = n^3 e^{-n}an​=n3e−n.

Answer: Convergent. L=(n+1)3n3⋅1e=1e<1L = \frac{(n+1)^3}{n^3} \cdot \frac{1}{e} = \frac{1}{e} < 1L=n3(n+1)3​⋅e1​=e1​<1.

Flashcard 3: Does an=n2(n+1)!a_n = \frac{n^2}{(n+1)!}an​=(n+1)!n2​ converge according to the Ratio Test?

Answer: Convergent. L=(n+1)2(n+2)!⋅(n+1)!n2=(n+1)2n2(n+2)→0<1L = \frac{(n+1)^2}{(n+2)!} \cdot \frac{(n+1)!}{n^2} = \frac{(n+1)^2}{n^2(n+2)} \to 0 < 1L=(n+2)!(n+1)2​⋅n2(n+1)!​=n2(n+2)(n+1)2​→0<1.

Flashcard 4: Evaluate convergence of an=2nn!a_n = \frac{2^n}{n!}an​=n!2n​ using the Ratio Test.

Answer: Convergent. L=2n+1→0<1L = \frac{2}{n+1} \to 0 < 1L=n+12​→0<1.

Flashcard 5: Determine the convergence of an=3n⋅n!a_n = 3^n \cdot n!an​=3n⋅n! using the Ratio Test.

Answer: Divergent. L=3(n+1)→∞>1L = 3(n+1) \to \infty > 1L=3(n+1)→∞>1.

Flashcard 6: Does an=n2na_n = \frac{n}{2^n}an​=2nn​ converge according to the Ratio Test?

Answer: Convergent. L=n+12n=12<1L = \frac{n+1}{2n} = \frac{1}{2} < 1L=2nn+1​=21​<1.

Flashcard 7: Does an=n23na_n = \frac{n^2}{3^n}an​=3nn2​ converge according to the Ratio Test?

Answer: Convergent. L=(n+1)23n2=13<1L = \frac{(n+1)^2}{3n^2} = \frac{1}{3} < 1L=3n2(n+1)2​=31​<1.

Flashcard 8: Identify the result of the Ratio Test for an=3nn!a_n = \frac{3^n}{n!}an​=n!3n​.

Answer: Convergent. L=3n+1→0<1L = \frac{3}{n+1} \to 0 < 1L=n+13​→0<1.

Flashcard 9: What is the convergence outcome for an=n3na_n = \frac{n}{3^n}an​=3nn​ using the Ratio Test?

Answer: Convergent. L=n+13n=13<1L = \frac{n+1}{3n} = \frac{1}{3} < 1L=3nn+1​=31​<1.

Flashcard 10: Evaluate an=n!2na_n = \frac{n!}{2^n}an​=2nn!​ using the Ratio Test. Convergent or divergent?

Answer: Divergent. L=n+12→∞>1L = \frac{n+1}{2} \to \infty > 1L=2n+1​→∞>1.

Flashcard 11: What does the Ratio Test say about an=1(n+1)!a_n = \frac{1}{(n+1)!}an​=(n+1)!1​?

Answer: Convergent. L=1n+2→0<1L = \frac{1}{n+2} \to 0 < 1L=n+21​→0<1.

Flashcard 12: State the Ratio Test conclusion for an=2nn!a_n = \frac{2^n}{n!}an​=n!2n​.

Answer: Convergent. L=2n+1→0<1L = \frac{2}{n+1} \to 0 < 1L=n+12​→0<1.

Flashcard 13: Evaluate an=n3n!a_n = \frac{n^3}{n!}an​=n!n3​ using the Ratio Test. Convergent or divergent?

Answer: Convergent. L=(n+1)3n3⋅1n+1=(n+1)2n3→0<1L = \frac{(n+1)^3}{n^3} \cdot \frac{1}{n+1} = \frac{(n+1)^2}{n^3} \to 0 < 1L=n3(n+1)3​⋅n+11​=n3(n+1)2​→0<1.

Flashcard 14: Identify the convergence of an=13na_n = \frac{1}{3^n}an​=3n1​ using the Ratio Test.

Answer: Convergent. L=13<1L = \frac{1}{3} < 1L=31​<1.

Flashcard 15: What does the Ratio Test conclude for an=n2e−na_n = n^2 e^{-n}an​=n2e−n?

Answer: Convergent. L=(n+1)2n2⋅1e=1e<1L = \frac{(n+1)^2}{n^2} \cdot \frac{1}{e} = \frac{1}{e} < 1L=n2(n+1)2​⋅e1​=e1​<1.

Flashcard 16: What is the result of Ratio Test for an=n!nna_n = \frac{n!}{n^n}an​=nnn!​?

Answer: Convergent. L=n+1n⋅nn(n+1)n+1=1e<1L = \frac{n+1}{n} \cdot \frac{n^n}{(n+1)^{n+1}} = \frac{1}{e} < 1L=nn+1​⋅(n+1)n+1nn​=e1​<1.

Flashcard 17: State the Ratio Test result for an=(n+1)!n!a_n = \frac{(n+1)!}{n!}an​=n!(n+1)!​.

Answer: Divergent. L=(n+2)!(n+1)!=n+2→∞>1L = \frac{(n+2)!}{(n+1)!} = n+2 \to \infty > 1L=(n+1)!(n+2)!​=n+2→∞>1.

Flashcard 18: What is the Ratio Test outcome for an=n!a_n = n!an​=n!?

Answer: Divergent. L=n+1→∞>1L = n+1 \to \infty > 1L=n+1→∞>1.

Flashcard 19: Is an=5nn!a_n = \frac{5^n}{n!}an​=n!5n​ convergent by the Ratio Test?

Answer: Convergent. L=5n+1→0<1L = \frac{5}{n+1} \to 0 < 1L=n+15​→0<1.

Flashcard 20: What condition must LLL satisfy for the series to diverge?

Answer: L>1L > 1L>1 or L=∞L = \inftyL=∞. When the ratio exceeds 1, terms grow without bound.

Flashcard 21: What condition must LLL satisfy for the series to converge?

Answer: L<1L < 1L<1. When the ratio is less than 1, terms shrink fast enough.

Flashcard 22: State the formula used in the Ratio Test for a series ana_nan​.

Answer: L=∣an+1∣∣an∣L = \frac{|a_{n+1}|}{|a_n|}L=∣an​∣∣an+1​∣​ as n→∞n \to \inftyn→∞. This is the limit of consecutive term ratios.

Flashcard 23: Evaluate convergence of an=2nn!a_n = \frac{2^n}{n!}an​=n!2n​ using the Ratio Test.

Answer: Convergent. L=2n+1→0<1L = \frac{2}{n+1} \to 0 < 1L=n+12​→0<1.

Flashcard 24: Is an=n!nna_n = \frac{n!}{n^n}an​=nnn!​ convergent by the Ratio Test?

Answer: Convergent. L=n+1n⋅nn(n+1)n+1=1e<1L = \frac{n+1}{n} \cdot \frac{n^n}{(n+1)^{n+1}} = \frac{1}{e} < 1L=nn+1​⋅(n+1)n+1nn​=e1​<1.

Flashcard 25: Apply the Ratio Test: an=n!3na_n = \frac{n!}{3^n}an​=3nn!​. Convergent or divergent?

Answer: Divergent. L=n+13→∞>1L = \frac{n+1}{3} \to \infty > 1L=3n+1​→∞>1.

Flashcard 26: Apply the Ratio Test: an=n2n!a_n = \frac{n^2}{n!}an​=n!n2​. Convergent or divergent?

Answer: Convergent. L=(n+1)2n2⋅1n+1=n+1n2→0<1L = \frac{(n+1)^2}{n^2} \cdot \frac{1}{n+1} = \frac{n+1}{n^2} \to 0 < 1L=n2(n+1)2​⋅n+11​=n2n+1​→0<1.

Flashcard 27: Apply the Ratio Test: an=5nn2a_n = \frac{5^n}{n^2}an​=n25n​. Convergent or divergent?

Answer: Divergent. L=5⋅n2(n+1)2=5>1L = 5 \cdot \frac{n^2}{(n+1)^2} = 5 > 1L=5⋅(n+1)2n2​=5>1.

Flashcard 28: State the limit test result for an=n22na_n = \frac{n^2}{2^n}an​=2nn2​ using the Ratio Test.

Answer: Convergent. L=(n+1)2n2⋅12=12<1L = \frac{(n+1)^2}{n^2} \cdot \frac{1}{2} = \frac{1}{2} < 1L=n2(n+1)2​⋅21​=21​<1.

Flashcard 29: Identify the series type: an=1n!a_n = \frac{1}{n!}an​=n!1​. Use the Ratio Test.

Answer: Convergent. L=1(n+1)→0<1L = \frac{1}{(n+1)} \to 0 < 1L=(n+1)1​→0<1.

Flashcard 30: State the Ratio Test result for an=(n+1)!n!a_n = \frac{(n+1)!}{n!}an​=n!(n+1)!​.

Answer: Divergent. L=(n+2)!(n+1)!=n+2→∞>1L = \frac{(n+2)!}{(n+1)!} = n+2 \to \infty > 1L=(n+1)!(n+2)!​=n+2→∞>1.