Explanation Given: The series is ∑ k = 0 ∞ a k = ∑ k = 0 ∞ 10 k 2 + 9 . Result used: (1) “Suppose f is a continuous, positive, decreasing function, for x ≥ 1 , and let a k = f ( k ) for k = 1 , 2 , 3 , ⋯ , then ∑ k = 1 ∞ a k and ∫ 1 ∞ f ( x ) d x either both converge or both diverge. (2) The function f ( x ) is decreasing function if f ′ ( x ) < 0 . Chain rule: d [ f ( x ) ] n d x = n [ f ( x ) ] n − 1 f ′ ( x ) Definition used: The improper integral ∫ a b f ( x ) d x is convergent if the corresponding limit exists. Calculation: Consider the function from given series f ( x ) = 10 x 2 + 9 . The derivative of the function is obtained as follows, f ′ ( x ) = 10 d d x ( x 2 + 9 ) − 1 = 10 ( − 1 ) ( x 2 + 9 ) − 1 − 1 d d x ( x 2 + 9 ) = − 10 ( x 2 + 9 ) 2 ( 2 x ) = − 20 x ( x 2 + 9 ) 2 Since f ′ ( x ) < 0 for x ≥ 0 , the given function is decreasing by using the Result (2). Clearly, the function f ′ ( x ) is continuous, positive and decreasing for x ≥ 0 . Use the Result (1), the given series is divergent if the improper integral ∫ 0 ∞ 10 x 2 + 9 d x is divergent

Single Variable Calculus: Early Transcendentals, Books a la Carte, and MyLab Math with Pearson eText -- Title-Specific Access Card Package (3rd Edition)

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ISBN: 9780134996103

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

Publisher: PEARSON

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Chapter 10.4, Problem 52E

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Chapter 10.4, Problem 51E

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Single Variable Calculus: Early Transcendentals, Books a la Carte, and MyLab Math with Pearson eText -- Title-Specific Access Card Package (3rd Edition)

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Whether the series convergence or divergence of the series or state that the test does not apply.

Solution Summary: The author explains that the series converges or diverges. The function f(x) is decreasing if the corresponding limit exists.

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Whether the series convergence or divergence of the series or state that the test does not apply.

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Chapter 10 Solutions