Explanation Given: The series is ∑ k = 0 ∞ k ! p k ( k + 1 ) k . Result used: Let the series ∑ a k of non-negative terms and r = lim k → ∞ | a k + 1 a k | . (i) If 0 ≤ r < 1 , the series converges. (ii) If r > 1 ( including r = ∞ ) , the series diverges. (iii) If r = 1 , the test is inconclusive Theorem used: The geometric series ∑ k = 0 ∞ a r k converges if | r | < 1 . Calculation: From the given series, the k th term is, a k = k ! p k ( k + 1 ) k . Thus, the ( k + 1 ) th term is, a k + 1 = ( k + 1 ) ! p k + 1 ( ( k + 1 ) + 1 ) k + 1 . Obtain the limit of | a k + 1 a k | . lim k → ∞ | a k + 1 a k | = lim k → ∞ | ( k + 1 ) ! p k + 1 ( ( k + 1 ) + 1 ) k + 1 k ! p k ( k + 1 ) k | = lim k → ∞ | ( k + 1 ) ! p k + 1 ( k + 1 ) k k ! p k ( k + 2 ) k + 1 | = lim k → ∞ k ! p ( k + 1 ) k + 1 k ! ( k + 2 ) k + 1 = lim k → ∞ p ( k + 1 k + 1 ) k + 1 That is, lim k → ∞ | a k + 1 a k | = lim k → ∞ p 1 ( 1 + 1 k + 1 ) k + 1 (1) Obtain the limit of ( 1 + 1 k + 1 ) k + 1

Calculus: Early Transcendentals (2nd Edition)

2nd Edition

ISBN: 9780321947345

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

Publisher: PEARSON

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Chapter 8.5, Problem 74E

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To find: The values of the parameter p>0 for the series converges.

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Chapter 8.5, Problem 73E

Chapter 8.5, Problem 75E

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Calculus: Early Transcendentals (2nd Edition)

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