Explanation Given: The given function ∫ k = 1 ∞ 6 k ( 3 k + 1 − 2 k + 1 ) ( 3 k − 2 k ) . Calculation: Alternating Series: An alternating series is an infinite series of the form ∑ n = 0 ∞ ( − 1 ) n a n or ∑ n = 0 ∞ ( − 1 ) n − 1 a n where a n > 0 ∀ n . Note: An alternating series converges if and only if the associated sequence of partial sum converges. Infinite Series: An infinite series is an expression of the form a 1 + a 2 + a 3 + ... = ∑ k = 1 ∞ a k and the n t h parallel sum of the series is provided by S n = a 1 + a 2 + a 3 + ... + a n = ∑ k − 1 n a k The series converge with sum S if the sequence of the partial sums { S n } converges to S . Thus, ∑ k = 1 n a k = lim n → ∞ S n That is, the sum defines to the limit of the sequence of partial sums converges. If the sequence { S n } does not converge, then the series ∑ k = 1 n a k diverges. The objective is to express ∫ k = 1 ∞ 6 k ( 3 k + 1 − 2 k + 1 ) ( 3 k − 2 k ) as a rational number. Let S ( n ) be the n t h partial sum of the given series. Then, S ( n ) = ∫ k = 1 ∞ 6 k ( 3 k + 1 − 2 k + 1 ) ( 3 k − 2 k ) = ∑ k = 1 n 6 k ⋅ 1 ( 3 k + 1 − 2 k + 1 ) ( 3 k − 2 k ) =

Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

6th Edition

ISBN: 9781524908102

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 8, Problem 97SP

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Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

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To express: the given function asa rational number.

Solution Summary: The author explains how the given function is expressed as a rational number. An alternating series is an expression of the form displaystyle

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Chapter 8 Solutions