AP Calculus BC Flashcards: Alternating Series Error Bound

What this deck covers

This deck focuses on Alternating Series Error Bound, 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.

Flashcard 1: What condition must $a_n$ satisfy for the Alternating Series Test?

Answer: $a_n$ must decrease and limit to zero. Both decreasing and limiting to zero are required.

Flashcard 2: What ensures the accuracy of an alternating series sum?

Answer: A decreasing error bound ensures accuracy. Small error bounds mean the partial sum is close to the true sum.

Flashcard 3: Evaluate if $\sum (-1)^n/n!$ satisfies the Alternating Series Test.

Answer: Yes, terms decrease and limit to zero; it converges. $\frac{1}{n!}$ decreases rapidly and approaches 0.

Flashcard 4: Estimate the error bound for the 6th partial sum of $\sum (-1)^n/n^6$.

Answer: Error is less than or equal to $\frac{1}{7^6}$. The 7th term is $\frac{1}{7^6}$ for the 6th partial sum.

Flashcard 5: What is the importance of the error bound in an alternating series?

Answer: It estimates the maximum error in the sum approximation. It provides a bound on how close the partial sum is to the true sum.

Flashcard 6: Determine if $\sum (-1)^n/(n^4 + n)$ converges.

Answer: Yes, it converges since terms decrease and limit to zero. $\frac{1}{n^4+n}$ decreases and approaches 0.

Flashcard 7: Estimate the error bound for the 5th partial sum of $\sum (-1)^n/n^4$.

Answer: Error is less than or equal to $\frac{1}{6^4} = \frac{1}{1296}$. The 6th term is $\frac{1}{6^4}$ for the 5th partial sum.

Flashcard 8: What is the effect of a larger $n$ on the error bound?

Answer: A larger $n$ decreases the error bound. Higher $n$ means the next term $|a_{n+1}|$ is smaller.

Flashcard 9: Why must $a_n$ decrease in an alternating series?

Answer: To ensure convergence and a valid error bound. Decreasing terms ensure the error bound formula works.

Flashcard 10: What does $|a_{n+1}|$ represent in the error bound formula?

Answer: It represents the absolute value of the next term. This is the magnitude of the first omitted term in the sum.

Flashcard 11: Identify the condition for an alternating series to converge.

Answer: Terms must decrease in absolute value and limit to zero. These are the two conditions for the Alternating Series Test.

Flashcard 12: How is the error bound expressed in terms of $n$?

Answer: Error $E_n \text{ is less than or equal to } |a_{n+1}|$. The error is bounded by the absolute value of the next term.

Flashcard 13: Calculate the error bound for $\sum (-1)^n/(n^2 + 1)$ at $n=4$.

Answer: Error is less than or equal to $\frac{1}{5^2 + 1} = \frac{1}{26}$. The 5th term is $\frac{1}{5^2+1}$ for the 4th partial sum.

Flashcard 14: What happens if $a_n$ does not decrease?

Answer: The series may not converge; error bound invalid. Without decreasing terms, the alternating series test fails.

Flashcard 15: Explain the significance of $a_n \to 0$ in the Alternating Series Test.

Answer: It ensures terms decrease to zero, aiding convergence. Without this limit, the series cannot converge.

Flashcard 16: Find the error bound for the 4th partial sum of $\sum (-1)^n/n^3$.

Answer: Error is less than or equal to $\frac{1}{5^3} = \frac{1}{125}$. The 5th term gives the error bound for the 4th partial sum.

Flashcard 17: What is the role of the next term in the error bound?

Answer: The next term's absolute value sets the error limit. It provides the upper bound for the approximation error.

Flashcard 18: Find the error bound for the 3rd partial sum of $\sum (-1)^n/n^5$.

Answer: Error is less than or equal to $\frac{1}{4^5} = \frac{1}{1024}$. The 4th term is $\frac{1}{4^5}$ for the 3rd partial sum.

Flashcard 19: Why does the series $\sum (-1)^n/n^3$ converge?

Answer: Terms decrease and limit to zero. $\frac{1}{n^3}$ decreases and approaches 0 as $n$ increases.

Flashcard 20: Determine if $\sum (-1)^n/(n+1)$ satisfies the conditions of the test.

Answer: Yes, terms decrease and limit to zero; it converges. $\frac{1}{n+1}$ decreases and approaches 0.

Flashcard 21: What happens to the error as more terms are added?

Answer: The error generally decreases. More terms provide better approximations with smaller errors.

Flashcard 22: How does the error bound relate to convergence?

Answer: A decreasing error bound indicates convergence. Smaller errors indicate the partial sums approach the true sum.

Flashcard 23: What condition must $a_n$ satisfy for the Alternating Series Test?

Answer: $a_n$ must decrease and limit to zero. Both decreasing and limiting to zero are required.

Flashcard 24: Predict the convergence of $\sum (-1)^n/(2^n)$.

Answer: Converges, terms decrease and limit to zero. $\frac{1}{2^n}$ decreases geometrically to 0.

Flashcard 25: Does the series $\sum (-1)^n n$ meet the Alternating Series Test?

Answer: No, terms do not decrease to zero. The terms $n$ increase without bound, violating the test.

Flashcard 26: State the conclusion of the Alternating Series Test.

Answer: The series converges if conditions are met. If both conditions hold, the alternating series converges.

Flashcard 27: Can the error bound be used for any series?

Answer: No, only for convergent alternating series. The series must be alternating and satisfy the test conditions.

Flashcard 28: What is the maximum possible error for an alternating series approximation?

Answer: Error $\text{E}_n = |a_{n+1}|$. The error equals the absolute value of the next unused term.

Flashcard 29: What is the formula for the Alternating Series Error Bound?

Answer: Error $\text{E}_n \text{ is less than or equal to } |a_{n+1}|$. The error is bounded by the absolute value of the next term.

Flashcard 30: What is the Alternating Series Test?

Answer: A series converges if terms decrease in absolute value and limit to zero. This is Leibniz's test for alternating series convergence.