Home

Tutoring

Subjects

Live Classes

Study Coach

Essay Review

On-Demand Courses

Colleges

Games

Opening subject page...

Loading your content

  1. My Subjects
  2. AP Calculus BC
  3. Flashcards

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.

← Back to flashcard decks

What this deck covers

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.

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: Alternating Series Test For Convergence

1

/ 30

0 reviewed

0% Complete

0 reviewing
QUESTION

Identify if ∑n=1∞(−1)n(1+1n)\textstyle\sum_{n=1}^{\infty} (-1)^n (1 + \frac{1}{n})∑n=1∞​(−1)n(1+n1​) converges.

Tap or drag to reveal answer

ANSWER

No, it does not satisfy an→0a_n \to 0an​→0 as n→∞n \to \inftyn→∞. The limit is 111, not 000, so the series diverges.

Swipe Right = I Know It! 🎉

Swipe Left = Still Learning

All flashcards

Flashcard 1: Identify if ∑n=1∞(−1)n(1+1n)\textstyle\sum_{n=1}^{\infty} (-1)^n (1 + \frac{1}{n})∑n=1∞​(−1)n(1+n1​) converges.

Answer: No, it does not satisfy an→0a_n \to 0an​→0 as n→∞n \to \inftyn→∞. The limit is 111, not 000, so the series diverges.

Flashcard 2: Does the series \bigsumn=1inf(−1)n1n\textstyle\bigsum_{n=1}^{\text{inf}} (-1)^n \frac{1}{n}\bigsumn=1inf​(−1)nn1​ satisfy an→0a_n \to 0an​→0?

Answer: Yes, an=1n→0a_n = \frac{1}{n} \to 0an​=n1​→0 as n→infn \to \text{inf}n→inf. The harmonic terms 1n\frac{1}{n}n1​ clearly approach zero as n→∞n \to \inftyn→∞.

Flashcard 3: 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 4: What is the condition for the terms ana_nan​ to be considered decreasing?

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

Flashcard 5: Find the limit of ana_nan​ for \bigsumn=1inf(−1)n1n\textstyle\bigsum_{n=1}^{\text{inf}} (-1)^n \frac{1}{n}\bigsumn=1inf​(−1)nn1​.

Answer: Limit an=1n→0a_n = \frac{1}{n} \to 0an​=n1​→0 as n→infn \to \text{inf}n→inf. As nnn increases, 1n\frac{1}{n}n1​ approaches zero.

Flashcard 6: Determine if the series ∑n=1∞(−1)n1n3\sum_{n=1}^{\infty} (-1)^n \frac{1}{n^3}∑n=1∞​(−1)nn31​ converges.

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

Flashcard 7: State the limit condition for the Alternating Series Test.

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

Flashcard 8: Find if ∑n=1∞(−1)n1ln⁡(n+1)\textstyle\sum_{n=1}^{\infty} (-1)^n \frac{1}{\ln(n+1)}∑n=1∞​(−1)nln(n+1)1​ converges.

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

Flashcard 9: Does the series \bigsumn=1inf(−1)n1n2\textstyle\bigsum_{n=1}^{\text{inf}} (-1)^n \frac{1}{n^2}\bigsumn=1inf​(−1)nn21​ converge?

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

Flashcard 10: 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 11: What does it mean for the terms to be 'eventually decreasing'?

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

Flashcard 12: 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 13: 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 14: What happens if ana_nan​ does not decrease to 0 in an alternating series?

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

Flashcard 15: What is the result if an→0a_n \to 0an​→0 is not satisfied?

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

Flashcard 16: Identify the series that does not converge: (A) ∑(−1)n1n3\textstyle\sum (-1)^n \frac{1}{n^3}∑(−1)nn31​, (B) ∑(−1)nn\textstyle\sum (-1)^n n∑(−1)nn.

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

Flashcard 17: Determine whether ∑n=1∞(−1)nn\textstyle \sum_{n=1}^{\infty} (-1)^n n∑n=1∞​(−1)nn converges.

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

Flashcard 18: Identify the condition required for terms in the Alternating Series.

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

Flashcard 19: Find if ∑n=1∞(−1)n1n4\sum_{n=1}^{\infty} (-1)^n \frac{1}{n^4}∑n=1∞​(−1)nn41​ converges.

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

Flashcard 20: Does the series ∑n=1∞(−1)n(1+1n2)\textstyle \sum_{n=1}^{\infty} (-1)^n (1 + \frac{1}{n^2})∑n=1∞​(−1)n(1+n21​) converge?

Answer: No, it does not converge. The limit lim⁡n→∞(1+1n2)=1≠0\lim_{n\to\infty} (1 + \frac{1}{n^2}) = 1 \neq 0limn→∞​(1+n21​)=1=0.

Flashcard 21: Identify the series that converges: (A) ∑(−1)n1n2\sum (-1)^n \frac{1}{n^2}∑(−1)nn21​, (B) ∑(−1)nn\sum (-1)^n n∑(−1)nn.

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

Flashcard 22: Find if the series ∑n=1∞(−1)nnn+1\sum_{n=1}^{\infty} (-1)^n \frac{n}{n+1}∑n=1∞​(−1)nn+1n​ converges.

Answer: No, it does not converge. The limit lim⁡n→∞nn+1=1≠0\lim_{n\to\infty} \frac{n}{n+1} = 1 \neq 0limn→∞​n+1n​=1=0.

Flashcard 23: What is the error bound for an alternating series?

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

Flashcard 24: What type of series does the Alternating Series Test apply to?

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

Flashcard 25: 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 26: 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 27: Find if the series ∑n=1∞(−1)n1n\sum_{n=1}^{\infty} (-1)^n \frac{1}{n}∑n=1∞​(−1)nn1​ satisfies decreasing terms.

Answer: Yes, 1n+1<1n\frac{1}{n+1} < \frac{1}{n}n+11​<n1​ for n→∞n \to \inftyn→∞. Since n+1>nn+1 > nn+1>n, the reciprocals decrease monotonically.

Flashcard 28: Does the series ∑n=1∞(−1)n1n2\sum_{n=1}^{\infty} (-1)^n \frac{1}{n^2}∑n=1∞​(−1)nn21​ converge?

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

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

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

Flashcard 30: Find if \bigsumn=1inf(−1)n1ln(n+1)\textstyle\bigsum_{n=1}^{\text{inf}} (-1)^n \frac{1}{\text{ln}(n+1)}\bigsumn=1inf​(−1)nln(n+1)1​ converges.

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