Explanation Given: The alternating series is ∑ k = 0 ∞ ( − 1 ) k 3 k + 1 = π 3 9 + ln 2 3 . 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 is ∑ k = 0 ∞ ( − 1 ) k 3 k + 1 . ∑ k = 0 ∞ ( − 1 ) k 3 k + 1 = 1 3 ( 0 ) + 1 − 1 ( 3 ( 1 ) + 1 ) + 1 ( 3 ( 2 ) + 1 ) − 1 ( 3 ( 3 ) + 1 ) + ⋯ + ( − 1 ) n ( 3 n + 1 ) + (

Student Solutions Manual, Single Variable for Calculus: Early Transcendentals

2nd Edition

ISBN: 9780321954329

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

Publisher: PEARSON

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Chapter 8.6, Problem 35E

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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 34E

Chapter 8.6, Problem 36E

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Chapter 8 Solutions

Student Solutions Manual, Single Variable for Calculus: Early Transcendentals

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