Explanation The given infinite series is, ∑ n = 1 ∞ 2 n + 1 n 2 ( n + 1 ) 2 . Let the partial sum S n be the sum of first n terms of the series, that is, S n = ∑ k = 1 n 2 k + 1 k 2 ( k + 1 ) 2 Obtain the first five partial sums of the above series as follows. S 1 = ∑ k = 1 1 2 k + 1 k 2 ( k + 1 ) 2 = 2 ( 1 ) + 1 ( 1 ) 2 ( 1 + 1 ) 2 = 3 4 = 0.75 When n = 2 the partial sum becomes, S 2 = ∑ k = 1 2 2 k + 1 k 2 ( k + 1 ) 2 = 2 ( 1 ) + 1 ( 1 ) 2 ( 1 + 1 ) 2 + 2 ( 2 ) + 1 ( 2 ) 2 ( 2 + 1 ) 2 = 3 4 + 5 36 = 32 36 That is, S 2 ≈ 0.89 . Since the calculation is little bit stuff without calculator, so use graphing calculation to compute the remaining partial sums. From the above screenshot, the partial sum S 3 can be calculated as, S 3 = ∑ k = 1 3 2 k + 1 k 2 ( k + 1 ) 2 = 2 ( 1 ) + 1 ( 1 ) 2 ( 1 + 1 ) 2 + 2 ( 2 ) + 1 ( 2 ) 2 ( 2 + 1 ) 2 + 2 ( 3 ) + 1 ( 3 ) 2 ( 3 + 1 ) 2 = 3 4 + 5 36 + 7 144 ≈ 0

Pearson eText for Basic Technical Mathematics with Calculus -- Instant Access (Pearson+)

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ISBN: 9780137554843

Author: Allyn Washington, Richard Evans

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Question

Chapter 30.1, Problem 21E

To determine

The first five partial sums of the given series and determine whether the series is convergent or not, if it so find its sum.

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Chapter 30.1, Problem 20E

Chapter 30.1, Problem 22E

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Chapter 30 Solutions

Pearson eText for Basic Technical Mathematics with Calculus -- Instant Access (Pearson+)

Ch. 30.1 - Prob. 1PE Ch. 30.1 - Prob. 2PE Ch. 30.1 - Prob. 1E Ch. 30.1 - Prob. 2E Ch. 30.1 - Prob. 3E Ch. 30.1 - Prob. 4E Ch. 30.1 - Prob. 5E Ch. 30.1 - Prob. 6E Ch. 30.1 - Prob. 7E Ch. 30.1 - Prob. 8E

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