Explanation Given: The alternating series is ∑ k = 1 ∞ ( − 1 ) k + 1 k 6 = 31 π 6 30 , 240 . Result used: Remainder Alternating Series “Let ∑ ( − 1 ) k + 1 a k be a convergent alternating series with terms that are non-increasing in magnitude. Let R n = S − S n be the remainder in approximating the value of the series by the sum of its first n terms. Then | R n | ≤ a n + 1 ”. Calculation: The given series ∑ k = 1 ∞ ( − 1 ) k + 1 k 6 . ∑ k = 1 ∞ ( − 1 ) k + 1 k 6 = − 1 1 6 + 1 2 6 − 1 3 6 ⋯ + ( − 1 ) n n 6 + ( − 1 ) n + 1 (

Calculus: Early Transcendentals, Books a la Carte Plus MyLab Math/MyLab Statistics Student Access Kit (2nd Edition)

2nd Edition

ISBN: 9780321977298

Author: William L. Briggs, Lyle Cochran, Bernard Gillett

Publisher: PEARSON

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Chapter 8.6, Problem 36E

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The number of terms added to the convergent series that the remainder is less than 10−4.

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Chapter 8.6, Problem 35E

Chapter 8.6, Problem 37E

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Chapter 8 Solutions

Calculus: Early Transcendentals, Books a la Carte Plus MyLab Math/MyLab Statistics Student Access Kit (2nd Edition)

Ch. 8.1 - Define sequence and give an example.Ch. 8.1 - Suppose the sequence {an} is defined by the...Ch. 8.1 - Suppose the sequence {an} is defined by the...Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E Ch. 8.1 - Given the series k=1k, evaluate the first four...Ch. 8.1 - The terms of a sequence of partial sums are...Ch. 8.1 - Consider the infinite series k=11k. Evaluate the...Ch. 8.1 - Explicit formulas Write the first four terms of...Ch. 8.1 - Explicit formulas Write the first four terms of...

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The number of terms added to the convergent series that the remainder is less than 10 − 4 .

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The number of terms added to the convergent series that the remainder is less than 10−4.

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Chapter 8 Solutions