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.

AP Calculus BC

Ratio Test For Convergence

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QUESTION
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Does an=n2na_n = \frac{n}{2^n} converge according to the Ratio Test?

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ANSWER

Convergent. L=n+12n=12<1L = \frac{n+1}{2n} = \frac{1}{2} < 1.

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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.

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Flashcard 1: Does an=n2na_n = \frac{n}{2^n} converge according to the Ratio Test?

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

Flashcard 2: Apply the Ratio Test: an=n!3na_n = \frac{n!}{3^n}. Convergent or divergent?

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

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

Answer: Convergent. L=(n+1)3n31n+1=(n+1)2n30<1L = \frac{(n+1)^3}{n^3} \cdot \frac{1}{n+1} = \frac{(n+1)^2}{n^3} \to 0 < 1.

Flashcard 4: What is the convergence outcome for an=n3na_n = \frac{n}{3^n} using the Ratio Test?

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

Flashcard 5: Is an=12na_n = \frac{1}{2^n} convergent by the Ratio Test?

Answer: Convergent. L=12<1L = \frac{1}{2} < 1.

Flashcard 6: What does the Ratio Test say about an=1(n+1)!a_n = \frac{1}{(n+1)!}?

Answer: Convergent. L=1n+20<1L = \frac{1}{n+2} \to 0 < 1.

Flashcard 7: Determine the convergence of an=3nn!a_n = 3^n \cdot n! using the Ratio Test.

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

Flashcard 8: Is an=n!nna_n = \frac{n!}{n^n} convergent by the Ratio Test?

Answer: Convergent. L=n+1nnn(n+1)n+1=1e<1L = \frac{n+1}{n} \cdot \frac{n^n}{(n+1)^{n+1}} = \frac{1}{e} < 1.

Flashcard 9: Evaluate an=n!2na_n = \frac{n!}{2^n} using the Ratio Test. Convergent or divergent?

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

Flashcard 10: Is an=n!nna_n = \frac{n!}{n^n} convergent by the Ratio Test?

Answer: Convergent. L=n+1nnn(n+1)n+1=1e<1L = \frac{n+1}{n} \cdot \frac{n^n}{(n+1)^{n+1}} = \frac{1}{e} < 1.

Flashcard 11: State the Ratio Test result for an=(n+1)!n!a_n = \frac{(n+1)!}{n!}.

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

Flashcard 12: What does the Ratio Test conclude for an=n2ena_n = n^2 e^{-n}?

Answer: Convergent. L=(n+1)2n21e=1e<1L = \frac{(n+1)^2}{n^2} \cdot \frac{1}{e} = \frac{1}{e} < 1.

Flashcard 13: Determine the Ratio Test result for an=n3ena_n = n^3 e^{-n}.

Answer: Convergent. L=(n+1)3n31e=1e<1L = \frac{(n+1)^3}{n^3} \cdot \frac{1}{e} = \frac{1}{e} < 1.

Flashcard 14: Is an=5nn!a_n = \frac{5^n}{n!} convergent by the Ratio Test?

Answer: Convergent. L=5n+10<1L = \frac{5}{n+1} \to 0 < 1.

Flashcard 15: State the limit test result for an=n22na_n = \frac{n^2}{2^n} using the Ratio Test.

Answer: Convergent. L=(n+1)2n212=12<1L = \frac{(n+1)^2}{n^2} \cdot \frac{1}{2} = \frac{1}{2} < 1.

Flashcard 16: Evaluate an=n!2na_n = \frac{n!}{2^n} using the Ratio Test. Convergent or divergent?

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

Flashcard 17: What does the Ratio Test conclude for an=n2ena_n = n^2 e^{-n}?

Answer: Convergent. L=(n+1)2n21e=1e<1L = \frac{(n+1)^2}{n^2} \cdot \frac{1}{e} = \frac{1}{e} < 1.

Flashcard 18: Determine the convergence of an=3nn!a_n = 3^n \cdot n! using the Ratio Test.

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

Flashcard 19: Identify the series type: an=1n!a_n = \frac{1}{n!}. Use the Ratio Test.

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

Flashcard 20: State the Ratio Test conclusion for an=2nn!a_n = \frac{2^n}{n!}.

Answer: Convergent. L=2n+10<1L = \frac{2}{n+1} \to 0 < 1.

Flashcard 21: State the formula used in the Ratio Test for a series ana_n.

Answer: L=an+1anL = \frac{|a_{n+1}|}{|a_n|} as nn \to \infty. This is the limit of consecutive term ratios.

Flashcard 22: What is the result of Ratio Test for an=n!nna_n = \frac{n!}{n^n}?

Answer: Convergent. L=n+1nnn(n+1)n+1=1e<1L = \frac{n+1}{n} \cdot \frac{n^n}{(n+1)^{n+1}} = \frac{1}{e} < 1.

Flashcard 23: Determine the Ratio Test result for an=n3ena_n = n^3 e^{-n}.

Answer: Convergent. L=(n+1)3n31e=1e<1L = \frac{(n+1)^3}{n^3} \cdot \frac{1}{e} = \frac{1}{e} < 1.

Flashcard 24: State the Ratio Test conclusion for an=2nn!a_n = \frac{2^n}{n!}.

Answer: Convergent. L=2n+10<1L = \frac{2}{n+1} \to 0 < 1.

Flashcard 25: What condition must LL satisfy for the series to diverge?

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

Flashcard 26: Identify the result of the Ratio Test for an=3nn!a_n = \frac{3^n}{n!}.

Answer: Convergent. L=3n+10<1L = \frac{3}{n+1} \to 0 < 1.

Flashcard 27: Does an=n2na_n = \frac{n}{2^n} converge according to the Ratio Test?

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

Flashcard 28: Apply the Ratio Test: an=n!3na_n = \frac{n!}{3^n}. Convergent or divergent?

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

Flashcard 29: Does an=n2(n+1)!a_n = \frac{n^2}{(n+1)!} 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 < 1.

Flashcard 30: Does an=n23na_n = \frac{n^2}{3^n} converge according to the Ratio Test?

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

Flashcard 31: Does an=n2(n+1)!a_n = \frac{n^2}{(n+1)!} 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 < 1.

Flashcard 32: Identify the convergence of an=13na_n = \frac{1}{3^n} using the Ratio Test.

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

Flashcard 33: What does the Ratio Test say about an=1(n+1)!a_n = \frac{1}{(n+1)!}?

Answer: Convergent. L=1n+20<1L = \frac{1}{n+2} \to 0 < 1.

Flashcard 34: Evaluate an=n3n!a_n = \frac{n^3}{n!} using the Ratio Test. Convergent or divergent?

Answer: Convergent. L=(n+1)3n31n+1=(n+1)2n30<1L = \frac{(n+1)^3}{n^3} \cdot \frac{1}{n+1} = \frac{(n+1)^2}{n^3} \to 0 < 1.

Flashcard 35: Is an=5nn!a_n = \frac{5^n}{n!} convergent by the Ratio Test?

Answer: Convergent. L=5n+10<1L = \frac{5}{n+1} \to 0 < 1.

Flashcard 36: Evaluate convergence of an=2nn!a_n = \frac{2^n}{n!} using the Ratio Test.

Answer: Convergent. L=2n+10<1L = \frac{2}{n+1} \to 0 < 1.

Flashcard 37: Apply the Ratio Test: an=5nn2a_n = \frac{5^n}{n^2}. Convergent or divergent?

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

Flashcard 38: What condition must LL satisfy for the series to diverge?

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

Flashcard 39: What does L=1L = 1 imply about the convergence of the series?

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

Flashcard 40: What is the result of Ratio Test for an=n!nna_n = \frac{n!}{n^n}?

Answer: Convergent. L=n+1nnn(n+1)n+1=1e<1L = \frac{n+1}{n} \cdot \frac{n^n}{(n+1)^{n+1}} = \frac{1}{e} < 1.

Flashcard 41: What is the Ratio Test outcome for an=n!a_n = n!?

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

Flashcard 42: State the limit test result for an=n22na_n = \frac{n^2}{2^n} using the Ratio Test.

Answer: Convergent. L=(n+1)2n212=12<1L = \frac{(n+1)^2}{n^2} \cdot \frac{1}{2} = \frac{1}{2} < 1.

Flashcard 43: Evaluate convergence of an=2nn!a_n = \frac{2^n}{n!} using the Ratio Test.

Answer: Convergent. L=2n+10<1L = \frac{2}{n+1} \to 0 < 1.

Flashcard 44: What does L=1L = 1 imply about the convergence of the series?

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

Flashcard 45: Apply the Ratio Test: an=n2n!a_n = \frac{n^2}{n!}. Convergent or divergent?

Answer: Convergent. L=(n+1)2n21n+1=n+1n20<1L = \frac{(n+1)^2}{n^2} \cdot \frac{1}{n+1} = \frac{n+1}{n^2} \to 0 < 1.

Flashcard 46: Is an=12na_n = \frac{1}{2^n} convergent by the Ratio Test?

Answer: Convergent. L=12<1L = \frac{1}{2} < 1.

Flashcard 47: What condition must LL satisfy for the series to converge?

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

Flashcard 48: Does an=n23na_n = \frac{n^2}{3^n} converge according to the Ratio Test?

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

Flashcard 49: Identify the series type: an=1n!a_n = \frac{1}{n!}. Use the Ratio Test.

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

Flashcard 50: What is the Ratio Test outcome for an=n!a_n = n!?

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

Flashcard 51: Identify the result of the Ratio Test for an=3nn!a_n = \frac{3^n}{n!}.

Answer: Convergent. L=3n+10<1L = \frac{3}{n+1} \to 0 < 1.

Flashcard 52: State the Ratio Test result for an=(n+1)!n!a_n = \frac{(n+1)!}{n!}.

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

Flashcard 53: Apply the Ratio Test: an=n2n!a_n = \frac{n^2}{n!}. Convergent or divergent?

Answer: Convergent. L=(n+1)2n21n+1=n+1n20<1L = \frac{(n+1)^2}{n^2} \cdot \frac{1}{n+1} = \frac{n+1}{n^2} \to 0 < 1.

Flashcard 54: What condition must LL satisfy for the series to converge?

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

Flashcard 55: Apply the Ratio Test: an=5nn2a_n = \frac{5^n}{n^2}. Convergent or divergent?

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

Flashcard 56: What is the convergence outcome for an=n3na_n = \frac{n}{3^n} using the Ratio Test?

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

Flashcard 57: Identify the convergence of an=13na_n = \frac{1}{3^n} using the Ratio Test.

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