Explanation Given: The celebrated theorem states that the sequence prime numbers { p k } satisfies lim k → ∞ p k k ln k = 1 . Result used: “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. Proof: Consider the function from given series 1 k ln k . The derivative of the function f ( x ) = x − 1 ( ln x ) − 1 is obtained as follows, f ′ ( x ) = ( x − 1 ) d d x ( ( ln x ) − 1 ) + ( ln x ) − 1 ⋅ d d x ( x − 1 ) = 1 x ⋅ ( − ( ln x ) − 2 d d x ( ln x ) ) + 1 ( ln x ) ⋅ ( ( − 1 ) x − 1 − 1 ) = − 1 x ( ln x ) 2 ⋅ 1 x + 1 ( ln x ) ⋅ ( − x − 2 ) = − 1 + ln x x 2 ( ln x ) 2 Clearly, the function f ′ ( x ) is continuous, positive and decreasing for x ≥ 2 . Use the Result (1), the series is divergent if the improper integral ∫ 2 ∞ 1 x ln x d x is divergent. By the definition, the improper integral is divergent if the limit does not exist. Compute ∫ 2 ∞ 1 x ln x d x as shown below. ∫ 2 ∞ 1 x ln x d x = lim t → ∞ [ ∫ 2 t 1 x ln x d x ] (1) Obtain ∫ 2 t 1 x ln x d x

Calculus: Early Transcendentals (2nd Edition)

2nd Edition

ISBN: 9780321947345

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

Publisher: PEARSON

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Chapter 8.4, Problem 64E

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To show: The series ∑k=2∞1klnk diverges, implies that the series ∑k=1∞1pk diverges.

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Chapter 8.4, Problem 63E

Chapter 8.4, Problem 65E

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Chapter 8 Solutions

Calculus: Early Transcendentals (2nd Edition)

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The series ∑ k = 2 ∞ 1 k ln k diverges, implies that the series ∑ k = 1 ∞ 1 p k diverges.

Solution Summary: The author explains the celebrated theorem, which states that the sequence prime numbers leftp_kright satisfies the series.

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To show: The series ∑k=2∞1klnk diverges, implies that the series ∑k=1∞1pk diverges.

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