AP Calculus BC Flashcards: Alternating Series Test For Convergence

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.

AP Calculus BC

Alternating Series Test For Convergence

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QUESTION
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Find if the series n=1(1)n1n5\textstyle \sum_{n=1}^{\infty} (-1)^n \frac{1}{n^5} converges.

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ANSWER

Yes, it converges by the Alternating Series Test. Higher powers guarantee all alternating series test conditions are satisfied.

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Flashcard 1: Find if the series n=1(1)n1n5\textstyle \sum_{n=1}^{\infty} (-1)^n \frac{1}{n^5} converges.

Answer: Yes, it converges by the Alternating Series Test. Higher powers guarantee all alternating series test conditions are satisfied.

Flashcard 2: What is the Alternating Series Estimation Theorem?

Answer: Provides error bound for partial sums of convergent series. It quantifies how close partial sums are to the series sum.

Flashcard 3: Determine whether n=1(1)nn\textstyle \sum_{n=1}^{\infty} (-1)^n n converges.

Answer: No, the terms do not approach zero. The terms an=na_n = n grow without bound, violating the limit condition.

Flashcard 4: Does the series n=1(1)n(1+1n2)\textstyle\sum_{n=1}^{\infty} (-1)^n (1 + \frac{1}{n^2}) converge?

Answer: No, it does not converge. The limit limn(1+1n2)=10\lim_{n\to\infty} (1 + \frac{1}{n^2}) = 1 \neq 0.

Flashcard 5: State the Alternating Series Remainder Theorem.

Answer: The remainder is less than the first unused term. This gives an upper bound on the approximation error.

Flashcard 6: Find if \bigsumn=1inf(1)n1ln(n+1)\textstyle\bigsum_{n=1}^{\text{inf}} (-1)^n \frac{1}{\text{ln}(n+1)} converges.

Answer: Yes, it converges by the Alternating Series Test. All conditions are met: positive terms, decreasing, limit to zero.

Flashcard 7: Does the series \bigsumn=1inf(1)n1n\textstyle\bigsum_{n=1}^{\text{inf}} (-1)^n \frac{1}{n} satisfy an0a_n \to 0?

Answer: Yes, an=1n0a_n = \frac{1}{n} \to 0 as ninfn \to \text{inf}. The harmonic terms 1n\frac{1}{n} clearly approach zero as nn \to \infty.

Flashcard 8: What happens if ana_n does not decrease to 0 in an alternating series?

Answer: The series diverges. The series fails to meet the necessary convergence conditions.

Flashcard 9: Identify the series that does not converge: (A) (1)n1n3\textstyle\sum (-1)^n \frac{1}{n^3}, (B) (1)nn\textstyle\sum (-1)^n n.

Answer: (B) does not converge. Series (B) has unbounded terms while (A) satisfies all conditions.

Flashcard 10: State the limit condition for the Alternating Series Test.

Answer: limnan=0\textstyle \lim_{n \to \infty} a_n = 0. The terms must approach zero for the series to have a chance at convergence.

Flashcard 11: Does the series \bigsumn=1inf(1)n1n2\textstyle\bigsum_{n=1}^{\text{inf}} (-1)^n \frac{1}{n^2} converge?

Answer: Yes, it converges by the Alternating Series Test. All three conditions are satisfied: positive, decreasing, limit to zero.

Flashcard 12: What does it mean for the terms to be 'eventually decreasing'?

Answer: There exists NN such that an+1<ana_{n+1} < a_n for n>Nn > N. The decreasing condition only needs to hold for sufficiently large nn.

Flashcard 13: Determine if \bigsumn=1(1)n1n+1\textstyle\bigsum_{n=1}^{\infty} (-1)^n \frac{1}{n+1} satisfies an0a_n \to 0.

Answer: Yes, an=1n+10a_n = \frac{1}{n+1} \to 0 as nn \to \infty. The shifted harmonic terms still approach zero as required.

Flashcard 14: What type of series does the Alternating Series Test apply to?

Answer: Applies to series with terms (1)nan(-1)^n a_n. The alternating factor (1)n(-1)^n creates the sign pattern.

Flashcard 15: What does it mean if a series is absolutely convergent?

Answer: The series an\sum |a_n| converges. This means the series of absolute values also converges.

Flashcard 16: What does it mean if a series is absolutely convergent?

Answer: The series \bigsuman\textstyle\bigsum |a_n| converges. This means the series of absolute values also converges.

Flashcard 17: State the limit condition for the Alternating Series Test.

Answer: \biglimninfan=0\textstyle\biglim_{n \to \text{inf}} a_n = 0. The terms must approach zero for the series to have a chance at convergence.

Flashcard 18: Find if the series n=1(1)n1n5\textstyle\sum_{n=1}^{\infty} (-1)^n \frac{1}{n^5} converges.

Answer: Yes, it converges by the Alternating Series Test. Higher powers guarantee all alternating series test conditions are satisfied.

Flashcard 19: Find if n=1(1)n1ln(n+1)\textstyle\sum_{n=1}^{\infty} (-1)^n \frac{1}{\ln(n+1)} converges.

Answer: Yes, it converges by the Alternating Series Test. All conditions are met: positive terms, decreasing, limit to zero.

Flashcard 20: Find the error bound for n=1(1)n1n2\textstyle \sum_{n=1}^{\infty} (-1)^n \frac{1}{n^2} at N=3N=3.

Answer: Error R3<a4=116|R_3| < |a_4| = \frac{1}{16}. The fourth term gives the error bound: 142=116\frac{1}{4^2} = \frac{1}{16}.

Flashcard 21: Does the series n=1(1)n(1+1n2)\textstyle \sum_{n=1}^{\infty} (-1)^n (1 + \frac{1}{n^2}) converge?

Answer: No, it does not converge. The limit limn(1+1n2)=10\lim_{n\to\infty} (1 + \frac{1}{n^2}) = 1 \neq 0.

Flashcard 22: What is the significance of the term (1)n(-1)^n in an alternating series?

Answer: It causes the series to alternate in sign. The (1)n(-1)^n factor creates the required alternating pattern.

Flashcard 23: What happens if ana_n does not decrease to 0 in an alternating series?

Answer: The series diverges. The series fails to meet the necessary convergence conditions.

Flashcard 24: Find the limit of ana_n for \bigsumn=1inf(1)n1n\textstyle\bigsum_{n=1}^{\text{inf}} (-1)^n \frac{1}{n}.

Answer: Limit an=1n0a_n = \frac{1}{n} \to 0 as ninfn \to \text{inf}. As nn increases, 1n\frac{1}{n} approaches zero.

Flashcard 25: Find if n=1(1)n1n4\sum_{n=1}^{\infty} (-1)^n \frac{1}{n^4} converges.

Answer: Yes, it converges by the Alternating Series Test. Even higher powers guarantee faster convergence to zero.

Flashcard 26: State the Alternating Series Remainder Theorem.

Answer: The remainder is less than the first unused term. This gives an upper bound on the approximation error.

Flashcard 27: What is the error bound for an alternating series?

Answer: Error RN<aN+1|R_N| < |a_{N+1}| for partial sum SNS_N. The error magnitude is bounded by the next term's absolute value.

Flashcard 28: Does the series \bigsumn=1inf(1)n1n\textstyle\bigsum_{n=1}^{\text{inf}} (-1)^n \frac{1}{n} satisfy an0a_n \to 0?

Answer: Yes, an=1n0a_n = \frac{1}{n} \to 0 as ninfn \to \text{inf}. The harmonic terms 1n\frac{1}{n} clearly approach zero as nn \to \infty.

Flashcard 29: Identify the condition required for terms in the Alternating Series.

Answer: The terms ana_n must be positive: an>0a_n > 0. This ensures the series doesn't have negative terms interfering with convergence.

Flashcard 30: Does the series n=1(1)n1n2\sum_{n=1}^{\infty} (-1)^n \frac{1}{n^2} converge?

Answer: Yes, it converges by the Alternating Series Test. All three conditions are satisfied: positive, decreasing, limit to zero.

Flashcard 31: What is the Alternating Series Estimation Theorem?

Answer: Provides error bound for partial sums of convergent series. It quantifies how close partial sums are to the series sum.

Flashcard 32: What does it mean for the terms to be 'eventually decreasing'?

Answer: There exists NN such that an+1<ana_{n+1} < a_n for n>Nn > N. The decreasing condition only needs to hold for sufficiently large nn.

Flashcard 33: Determine if n=1(1)n1n+1\sum_{n=1}^{\infty} (-1)^n \frac{1}{n+1} satisfies an0a_n \to 0.

Answer: Yes, an=1n+10a_n = \frac{1}{n+1} \to 0 as nn \to \infty. The shifted harmonic terms still approach zero as required.

Flashcard 34: Determine if the series n=1(1)n1n3\sum_{n=1}^{\infty} (-1)^n \frac{1}{n^3} converges.

Answer: Yes, it converges by the Alternating Series Test. Higher powers ensure faster convergence with all conditions satisfied.

Flashcard 35: Determine if the series n=1(1)n1n3\textstyle \sum_{n=1}^{\infty} (-1)^n \frac{1}{n^3} converges.

Answer: Yes, it converges by the Alternating Series Test. Higher powers ensure faster convergence with all conditions satisfied.

Flashcard 36: What type of series does the Alternating Series Test apply to?

Answer: Applies to series with terms (1)nan(-1)^n a_n. The alternating factor (1)n(-1)^n creates the sign pattern.

Flashcard 37: Identify the condition required for terms in the Alternating Series.

Answer: The terms ana_n must be positive: an>0a_n > 0. This ensures the series doesn't have negative terms interfering with convergence.

Flashcard 38: What is the condition for the terms ana_n to be considered decreasing?

Answer: an+1<ana_{n+1} < a_n for all nn. This ensures the terms form a monotonically decreasing sequence.

Flashcard 39: Identify if n=1(1)n(1+1n)\textstyle\sum_{n=1}^{\infty} (-1)^n (1 + \frac{1}{n}) converges.

Answer: No, it does not satisfy an0a_n \to 0 as nn \to \infty. The limit is 11, not 00, so the series diverges.

Flashcard 40: Find if the series n=1(1)n1n\sum_{n=1}^{\infty} (-1)^n \frac{1}{n} satisfies decreasing terms.

Answer: Yes, 1n+1<1n\frac{1}{n+1} < \frac{1}{n} for nn \to \infty. Since n+1>nn+1 > n, the reciprocals decrease monotonically.

Flashcard 41: What is the significance of the term (1)n(-1)^n in an alternating series?

Answer: It causes the series to alternate in sign. The (1)n(-1)^n factor creates the required alternating pattern.

Flashcard 42: Identify the series that converges: (A) \bigsum(1)n1n2\textstyle\bigsum (-1)^n \frac{1}{n^2}, (B) \bigsum(1)nn\textstyle\bigsum (-1)^n n.

Answer: (A) converges by the Alternating Series Test. Series (A) satisfies all conditions while (B) has terms that don't approach zero.

Flashcard 43: What is the result if an0a_n \to 0 is not satisfied?

Answer: The series diverges. Without the limit condition, the series cannot converge.

Flashcard 44: State the primary purpose of the Alternating Series Test.

Answer: To determine if an alternating series converges. It checks the three key conditions for alternating series convergence.

Flashcard 45: Identify the series that converges: (A) (1)n1n2\sum (-1)^n \frac{1}{n^2}, (B) (1)nn\sum (-1)^n n.

Answer: (A) converges by the Alternating Series Test. Series (A) satisfies all conditions while (B) has terms that don't approach zero.

Flashcard 46: Find if n=1(1)n1n4\textstyle \sum_{n=1}^{\infty} (-1)^n \frac{1}{n^4} converges.

Answer: Yes, it converges by the Alternating Series Test. Even higher powers guarantee faster convergence to zero.

Flashcard 47: What does the Alternating Series Test determine about a series?

Answer: Determines if an alternating series converges. It's the specific test for alternating series convergence criteria.

Flashcard 48: Find the error bound for n=1(1)n1n2\textstyle\sum_{n=1}^{\infty} (-1)^n \frac{1}{n^2} at N=3N=3.

Answer: Error R3<a4=116|R_3| < |a_4| = \frac{1}{16}. The fourth term gives the error bound: 142=116\frac{1}{4^2} = \frac{1}{16}.

Flashcard 49: Find if the series n=1(1)n1n\sum_{n=1}^{\infty} (-1)^n \frac{1}{n} satisfies decreasing terms.

Answer: Yes, 1n+1<1n\frac{1}{n+1} < \frac{1}{n} for nn \to \infty. Since n+1>nn+1 > n, the reciprocals decrease monotonically.

Flashcard 50: What does the Alternating Series Test determine about a series?

Answer: Determines if an alternating series converges. It's the specific test for alternating series convergence criteria.

Flashcard 51: What is the error bound for an alternating series?

Answer: Error RN<aN+1|R_N| < |a_{N+1}| for partial sum SNS_N. The error magnitude is bounded by the next term's absolute value.

Flashcard 52: What is the result if an0a_n \to 0 is not satisfied?

Answer: The series diverges. Without the limit condition, the series cannot converge.

Flashcard 53: Identify if n=1(1)n(1+1n)\textstyle\sum_{n=1}^{\infty} (-1)^n (1 + \frac{1}{n}) converges.

Answer: No, it does not satisfy an0a_n \to 0 as nn \to \infty. The limit is 11, not 00, so the series diverges.

Flashcard 54: Find the limit of ana_n for n=1(1)n1n\sum_{n=1}^{\infty} (-1)^n \frac{1}{n}.

Answer: Limit an=1n0a_n = \frac{1}{n} \to 0 as nn \to \infty. As nn increases, 1n\frac{1}{n} approaches zero.

Flashcard 55: State the primary purpose of the Alternating Series Test.

Answer: To determine if an alternating series converges. It checks the three key conditions for alternating series convergence.

Flashcard 56: Find if the series n=1(1)nnn+1\sum_{n=1}^{\infty} (-1)^n \frac{n}{n+1} converges.

Answer: No, it does not converge. The limit limnnn+1=10\lim_{n\to\infty} \frac{n}{n+1} = 1 \neq 0.

Flashcard 57: Identify the series that does not converge: (A) (1)n1n3\textstyle\sum (-1)^n \frac{1}{n^3}, (B) (1)nn\textstyle\sum (-1)^n n.

Answer: (B) does not converge. Series (B) has unbounded terms while (A) satisfies all conditions.

Flashcard 58: Find the error bound for n=1(1)n1n\textstyle\sum_{n=1}^{\infty} (-1)^n \frac{1}{n} at N=2N=2.

Answer: Error R2<a3=13|R_2| < |a_3| = \frac{1}{3}. The third term in the harmonic series: 13\frac{1}{3}.

Flashcard 59: Find the error bound for n=1(1)n1n\textstyle \sum_{n=1}^{\infty} (-1)^n \frac{1}{n} at N=2N=2.

Answer: Error R2<a3=13|R_2| < |a_3| = \frac{1}{3}. The third term in the harmonic series: 13\frac{1}{3}

Flashcard 60: What is the condition for the terms ana_n to be considered decreasing?

Answer: an+1<ana_{n+1} < a_n for all nn. This ensures the terms form a monotonically decreasing sequence.

Flashcard 61: Find if the series n=1(1)nnn+1\textstyle\sum_{n=1}^{\infty} (-1)^n \frac{n}{n+1} converges.

Answer: No, it does not converge. The limit limnnn+1=10\lim_{n\to\infty} \frac{n}{n+1} = 1 \neq 0.