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This deck focuses on Alternating Series Test For Convergence, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus BC.
Study Alternating Series 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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Find if the series ∑n=1∞(−1)nn51 converges.
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Yes, it converges by the Alternating Series Test. Higher powers guarantee all alternating series test conditions are satisfied.
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This deck focuses on Alternating Series Test For Convergence, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus BC.
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.
Answer: Yes, it converges by the Alternating Series Test. Higher powers guarantee all alternating series test conditions are satisfied.
Answer: Provides error bound for partial sums of convergent series. It quantifies how close partial sums are to the series sum.
Answer: No, the terms do not approach zero. The terms an=n grow without bound, violating the limit condition.
Answer: No, it does not converge. The limit limn→∞(1+n21)=1=0.
Answer: The remainder is less than the first unused term. This gives an upper bound on the approximation error.
Answer: Yes, it converges by the Alternating Series Test. All conditions are met: positive terms, decreasing, limit to zero.
Answer: Yes, an=n1→0 as n→inf. The harmonic terms n1 clearly approach zero as n→∞.
Answer: The series diverges. The series fails to meet the necessary convergence conditions.
Answer: (B) does not converge. Series (B) has unbounded terms while (A) satisfies all conditions.
Answer: limn→∞an=0. The terms must approach zero for the series to have a chance at convergence.
Answer: Yes, it converges by the Alternating Series Test. All three conditions are satisfied: positive, decreasing, limit to zero.
Answer: There exists N such that an+1<an for n>N. The decreasing condition only needs to hold for sufficiently large n.
Answer: Yes, an=n+11→0 as n→∞. The shifted harmonic terms still approach zero as required.
Answer: Applies to series with terms (−1)nan. The alternating factor (−1)n creates the sign pattern.
Answer: The series ∑∣an∣ converges. This means the series of absolute values also converges.
Answer: The series \bigsum∣an∣ converges. This means the series of absolute values also converges.
Answer: \biglimn→infan=0. The terms must approach zero for the series to have a chance at convergence.
Answer: Yes, it converges by the Alternating Series Test. Higher powers guarantee all alternating series test conditions are satisfied.
Answer: Yes, it converges by the Alternating Series Test. All conditions are met: positive terms, decreasing, limit to zero.
Answer: Error ∣R3∣<∣a4∣=161. The fourth term gives the error bound: 421=161.
Answer: No, it does not converge. The limit limn→∞(1+n21)=1=0.
Answer: It causes the series to alternate in sign. The (−1)n factor creates the required alternating pattern.
Answer: The series diverges. The series fails to meet the necessary convergence conditions.
Answer: Limit an=n1→0 as n→inf. As n increases, n1 approaches zero.
Answer: Yes, it converges by the Alternating Series Test. Even higher powers guarantee faster convergence to zero.
Answer: The remainder is less than the first unused term. This gives an upper bound on the approximation error.
Answer: Error ∣RN∣<∣aN+1∣ for partial sum SN. The error magnitude is bounded by the next term's absolute value.
Answer: Yes, an=n1→0 as n→inf. The harmonic terms n1 clearly approach zero as n→∞.
Answer: The terms an must be positive: an>0. This ensures the series doesn't have negative terms interfering with convergence.
Answer: Yes, it converges by the Alternating Series Test. All three conditions are satisfied: positive, decreasing, limit to zero.
Answer: Provides error bound for partial sums of convergent series. It quantifies how close partial sums are to the series sum.
Answer: There exists N such that an+1<an for n>N. The decreasing condition only needs to hold for sufficiently large n.
Answer: Yes, an=n+11→0 as n→∞. The shifted harmonic terms still approach zero as required.
Answer: Yes, it converges by the Alternating Series Test. Higher powers ensure faster convergence with all conditions satisfied.
Answer: Yes, it converges by the Alternating Series Test. Higher powers ensure faster convergence with all conditions satisfied.
Answer: Applies to series with terms (−1)nan. The alternating factor (−1)n creates the sign pattern.
Answer: The terms an must be positive: an>0. This ensures the series doesn't have negative terms interfering with convergence.
Answer: an+1<an for all n. This ensures the terms form a monotonically decreasing sequence.
Answer: No, it does not satisfy an→0 as n→∞. The limit is 1, not 0, so the series diverges.
Answer: Yes, n+11<n1 for n→∞. Since n+1>n, the reciprocals decrease monotonically.
Answer: It causes the series to alternate in sign. The (−1)n factor creates the required alternating pattern.
Answer: (A) converges by the Alternating Series Test. Series (A) satisfies all conditions while (B) has terms that don't approach zero.
Answer: The series diverges. Without the limit condition, the series cannot converge.
Answer: To determine if an alternating series converges. It checks the three key conditions for alternating series convergence.
Answer: (A) converges by the Alternating Series Test. Series (A) satisfies all conditions while (B) has terms that don't approach zero.
Answer: Yes, it converges by the Alternating Series Test. Even higher powers guarantee faster convergence to zero.
Answer: Determines if an alternating series converges. It's the specific test for alternating series convergence criteria.
Answer: Error ∣R3∣<∣a4∣=161. The fourth term gives the error bound: 421=161.
Answer: Yes, n+11<n1 for n→∞. Since n+1>n, the reciprocals decrease monotonically.
Answer: Determines if an alternating series converges. It's the specific test for alternating series convergence criteria.
Answer: Error ∣RN∣<∣aN+1∣ for partial sum SN. The error magnitude is bounded by the next term's absolute value.
Answer: The series diverges. Without the limit condition, the series cannot converge.
Answer: No, it does not satisfy an→0 as n→∞. The limit is 1, not 0, so the series diverges.
Answer: Limit an=n1→0 as n→∞. As n increases, n1 approaches zero.
Answer: To determine if an alternating series converges. It checks the three key conditions for alternating series convergence.
Answer: No, it does not converge. The limit limn→∞n+1n=1=0.
Answer: (B) does not converge. Series (B) has unbounded terms while (A) satisfies all conditions.
Answer: Error ∣R2∣<∣a3∣=31. The third term in the harmonic series: 31.
Answer: Error ∣R2∣<∣a3∣=31. The third term in the harmonic series: 31
Answer: an+1<an for all n. This ensures the terms form a monotonically decreasing sequence.
Answer: No, it does not converge. The limit limn→∞n+1n=1=0.