Explanation Given information: The infinite series is ∑ k = 1 ∞ 5 5 k − 1 . Explanation: Here, f ( x ) = 5 5 x − 1 is a positive and continuous for x ≥ 1 . f ( x ) = 5 5 x − 1 can be written as f ( x ) = 5 ( 5 x − 1 ) − 1 f ' ( x ) = − 25 ( 5 x − 1 ) 2 So, f ' ( x ) is negative and hence f ( x ) = 5 5 x − 1 is a decreasing function. Integral test: Let f ( x ) be continuous, decreasing, and positive for x ≥ 1 . Then, the infinite series ∑ k = 1 ∞ f ( k ) is convergent if the improper integral ∫ 1 ∞ f ( x ) d x is convergent, and the infinite series is divergent if the improper integral is divergent. By using the integral test, ∫ 1 ∞ 5 5 x − 1 d x = lim b → ∞ ∫ 1 b 5 5 x − 1 d x Now, solve improper integral by using substitution method as follows: Substitute 5 x − 1 = u ⇒ 5 d x = d u When x = 1 , u = 4 ; when x = b , u = 5 b − 1

12th Edition

ISBN: 9780137590810

Author: Larry J. Goldstein, David C. Lay, David I. Schneider, Nakhle H. Asmar, William Edward Tavernetti

Publisher: PEARSON

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Chapter 11.4, Problem 5E

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Theinfinite series ∑k=1∞55k−1 is convergent or divergent by using the integral test.

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Chapter 11.4, Problem 4E

Chapter 11.4, Problem 6E

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Chapter 11 Solutions

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Theinfinite series ∑ k = 1 ∞ 5 5 k − 1 is convergent or divergent by using the integral test.

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Theinfinite series ∑k=1∞55k−1 is convergent or divergent by using the integral test.

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Chapter 11 Solutions