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AP Calculus BC Flashcards: Power Series Radius Interval Of Convergence

Study Power Series Radius Interval Of 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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What this deck covers

This deck focuses on Power Series Radius Interval Of 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: Power Series Radius Interval Of Convergence

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QUESTION

Determine the interval of convergence for \bigsumn=0∞(x−3)n2n\textstyle\bigsum_{n=0}^{\text{∞}} \frac{(x-3)^n}{2^n}\bigsumn=0∞​2n(x−3)n​.

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ANSWER

(1,5)(1, 5)(1,5). Series ∑(x−3)n2n\sum \frac{(x-3)^n}{2^n}∑2n(x−3)n​ has center 3, radius 2; endpoints diverge.

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Flashcard 1: Determine the interval of convergence for \bigsumn=0∞(x−3)n2n\textstyle\bigsum_{n=0}^{\text{∞}} \frac{(x-3)^n}{2^n}\bigsumn=0∞​2n(x−3)n​.

Answer: (1,5)(1, 5)(1,5). Series ∑(x−3)n2n\sum \frac{(x-3)^n}{2^n}∑2n(x−3)n​ has center 3, radius 2; endpoints diverge.

Flashcard 2: Determine the interval of convergence for ∑n=1∞(−1)nxnn\textstyle\sum_{n=1}^{\infty} \frac{(-1)^n x^n}{n}∑n=1∞​n(−1)nxn​.

Answer: (−1,1](-1, 1](−1,1]. Alternating series converges at x=1x = 1x=1; diverges at x=−1x = -1x=−1.

Flashcard 3: Determine the convergence interval of ∑n=1∞xnn\textstyle\sum_{n=1}^{\infty} \frac{x^n}{n}∑n=1∞​nxn​.

Answer: (−1,1](-1, 1](−1,1]. Alternating harmonic-type series; converges at x=1x = 1x=1 by alternating series test.

Flashcard 4: State the formula for the root test applied to a series.

Answer: lim⁡n→∞∣an∣1/n<1\textstyle\lim_{n \to \infty} \big|a_n\big|^{1/n} < 1limn→∞​​an​​1/n<1 for convergence. Root test condition for series convergence.

Flashcard 5: Find the interval of convergence for \bigsumn=0∞xnn!\textstyle\bigsum_{n=0}^{\text{∞}} \frac{x^n}{n!}\bigsumn=0∞​n!xn​.

Answer: (−∞,∞)(-\text{∞}, \text{∞})(−∞,∞). Exponential function series; factorial dominates, giving infinite radius.

Flashcard 6: What is the radius of convergence if L=0L = 0L=0 in the ratio test?

Answer: R=∞R = \text{∞}R=∞. When limit is 0, radius becomes infinite.

Flashcard 7: Identify the radius of convergence for an=1n2a_n = \frac{1}{n^2}an​=n21​ using the ratio test.

Answer: R=∞R = \text{∞}R=∞. Quadratic decay gives infinite radius via ratio test.

Flashcard 8: Which test confirms divergence if the limit is greater than 1?

Answer: Ratio test or root test. Both tests indicate divergence when limit exceeds 1.

Flashcard 9: Identify the radius of convergence for \bigsumn=0∞2nxn\textstyle\bigsum_{n=0}^{\text{∞}} 2^n x^n\bigsumn=0∞​2nxn.

Answer: R=12R = \frac{1}{2}R=21​. Rewrite as ∑(2x)n\sum (2x)^n∑(2x)n; radius is 12\frac{1}{2}21​.

Flashcard 10: What does L<1L < 1L<1 indicate in the ratio test for a series?

Answer: Series converges absolutely. Ratio test shows absolute convergence when limit is less than 1.

Flashcard 11: What is the radius of convergence when L=1L = 1L=1 in the ratio test?

Answer: Test is inconclusive. When L=1L = 1L=1, ratio test gives no information about convergence.

Flashcard 12: Find the radius of convergence for \bigsumn=1∞xnn3\textstyle\bigsum_{n=1}^{\text{∞}} \frac{x^n}{n^3}\bigsumn=1∞​n3xn​.

Answer: R=∞R = \text{∞}R=∞. Cubic decay gives infinite radius via ratio test.

Flashcard 13: State the interval of convergence for a series with R=0R = 0R=0.

Answer: x = c\text{x = c}x = c only. Zero radius restricts convergence to the center point only.

Flashcard 14: What test can be used to determine the convergence of a power series?

Answer: Ratio test or root test. Standard convergence tests for determining series behavior.

Flashcard 15: What does L=1L = 1L=1 imply in the ratio test for convergence?

Answer: Test is inconclusive. Ratio test fails when limit equals 1; other methods needed.

Flashcard 16: Determine the interval of convergence for ∑n=1∞xnn\sum_{n=1}^{\infty} \frac{x^n}{n}∑n=1∞​nxn​ using the ratio test.

Answer: (−1,1](-1, 1](−1,1]. Series ∑xnn\sum \frac{x^n}{n}∑nxn​ converges at x=1x = 1x=1 by alternating series test.

Flashcard 17: Which condition indicates absolute convergence of a series?

Answer: If \bigsum∣an∣\textstyle\bigsum |a_n|\bigsum∣an​∣ converges. Absolute convergence means the series of absolute values converges.

Flashcard 18: Find the radius of convergence for \bigsumn=1∞xn\textstyle\bigsum_{n=1}^{\text{∞}} x^n\bigsumn=1∞​xn using the root test.

Answer: R=1R = 1R=1. Root test gives lim⁡∣x∣=∣x∣\lim |x| = |x|lim∣x∣=∣x∣; radius is 1.

Flashcard 19: Identify the radius of convergence for a geometric series ∑xn\sum x^n∑xn.

Answer: R=1R = 1R=1. Standard geometric series has radius 1.

Flashcard 20: State the formula for determining convergence using the root test.

Answer: lim⁡n→∞∣an∣1/n<1\textstyle \lim_{n \to \infty} \big| a_n \big|^{1/n} < 1limn→∞​​an​​1/n<1. Root test convergence criterion.

Flashcard 21: Determine the radius of convergence for ∑n=0∞(2x)nn!\sum_{n=0}^{\infty} \frac{(2x)^n}{n!}∑n=0∞​n!(2x)n​.

Answer: R=∞R = \inftyR=∞. Factor out constants; factorial dominates for infinite radius.

Flashcard 22: What condition implies divergence in the root test?

Answer: lim⁡n→∞∣an∣1/n>1\textstyle \lim_{n \to \infty} \big| a_n \big|^{1/n} > 1limn→∞​​an​​1/n>1. Root test divergence condition.

Flashcard 23: Find the radius of convergence of ∑n=1∞xn3n\textstyle\sum_{n=1}^{\infty} \frac{x^n}{3^n}∑n=1∞​3nxn​.

Answer: R=3R = 3R=3. Rewrite as ∑(x3)n\sum (\frac{x}{3})^n∑(3x​)n; geometric series with ratio 13\frac{1}{3}31​.

Flashcard 24: Identify the convergence interval for \bigsumn=0∞xn\textstyle\bigsum_{n=0}^{\text{∞}} x^n\bigsumn=0∞​xn using the ratio test.

Answer: (−1,1)(-1, 1)(−1,1). Standard geometric series; endpoints both diverge.

Flashcard 25: What is the radius of convergence if L=∞L = \text{∞}L=∞ in the ratio test?

Answer: R=0R = 0R=0. Infinite limit in ratio test means zero radius.

Flashcard 26: What does it mean if a power series diverges for x=c+Rx = c + Rx=c+R?

Answer: Endpoint x=c+Rx = c + Rx=c+R not in interval of convergence. Endpoint divergence excludes that point from the convergence interval.

Flashcard 27: Identify the radius of convergence if R=0R = 0R=0 for a series.

Answer: Series converges only at x=cx = cx=c. Zero radius means convergence only at the center point.

Flashcard 28: What does the radius of convergence R=∞R = \text{∞}R=∞ imply?

Answer: Series converges for all xxx. Infinite radius means convergence for every real number.

Flashcard 29: What is the interval of convergence for a power series?

Answer: Values of xxx where the series converges. The set of all xxx-values where the power series converges.

Flashcard 30: Find the radius of convergence for an=1n!a_n = \frac{1}{n!}an​=n!1​ using the ratio test.

Answer: R=∞R = \text{∞}R=∞, series converges for all xxx. Factorial growth dominates, making the limit 0, so R=10=∞R = \frac{1}{0} = \inftyR=01​=∞.