Explanation Given info. Given series, S = ∑ n = 1 ∞ ( − 1 ) n + 1 1 n 3 Concept used: CAS, calculate the sum S n . Note that, the error in the N t h partial sum is at most a N + 1 . Calculation: (a) Calculate the sums as follows. S 1 = 1 1 3 = 1 S 2 = 1 − 1 2 3 = 0.875 S 3 = 1 − 1 2 3 + 1 3 3 = 0.912037 S 4 = 1 1 2 3 + 1 3 3 − 1 4 3 = 0.896412 S 5 = 1 − 1 2 3 + 1 3 3 − 1 4 3 + 1 5 3 = 0.904412 S 6 = 1 − 1 2 3 + 1 3 3 − 1 4 3 + 1 5 3 − 1 6 3 = 0.899782 S 7 = 1 − 1 2 3 + 1 3 3 − 1 4 3 + 1 5 3 − 1 6 3 + 1 7 3 = 0.902698 S 8 = 1 − 1 2 3 + 1 3 3 − 1 4 3 + 1 5 3 − 1 6 3 + 1 7 3 − 1 8 3 = 0.900745 S 9 = 1 − 1 2 3 + 1 3 3 − 1 4 3 + 1 5 3 − 1 6 3 + 1 7 3 − 1 8 3 + 1 9 3 = 0...

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 11.4, Problem 11E

To determine

Let $S = \sum_{n = 1}^{\infty} {(- 1)}^{n + 1} . \frac{1}{n^{3}}$

(a) Calculate $S_{n}$ for $1 \leq n \leq 10$

(b) Use $S_{n} - | S_{n} - S_{n} | < b_{n + 1}$ to show that

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Chapter 11.4, Problem 10E

Chapter 11.4, Problem 12E

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Chapter 11 Solutions

Calculus - Standalone book

Ch. 11.1 - Prob. 1PQ Ch. 11.1 - Prob. 2PQ Ch. 11.1 - Prob. 3PQ Ch. 11.1 - Prob. 4PQ Ch. 11.1 - Prob. 5PQ Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E

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Let S = ∑ n = 1 ∞ ( − 1 ) n + 1 . 1 n 3 (a) Calculate S n for 1 ≤ n ≤ 10 (b) Use S n − | S n − S n | < b n + 1 to show that

Solution Summary: The author explains that the error in the N t h partial sum is at most a N + 1.

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Let $S = \sum_{n = 1}^{\infty} {(- 1)}^{n + 1} . \frac{1}{n^{3}}$

(a) Calculate $S_{n}$ for $1 \leq n \leq 10$

(b) Use $S_{n} - | S_{n} - S_{n} | < b_{n + 1}$ to show that

Want to see the full answer?

Chapter 11 Solutions

Let S = ∑ n = 1 ∞ ( − 1 ) n + 1 . 1 n 3 (a) Calculate S n for 1 ≤ n ≤ 10 (b) Use S n − | S n − S n | &lt; b n + 1 to show that

Solution Summary: The author explains that the error in the N t h partial sum is at most a N + 1.

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Let S=∑n=1∞(−1)n+1.1n3 (a) Calculate Sn for 1≤n≤10 (b) Use Sn−|Sn−Sn|<bn+1 to show that

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Chapter 11 Solutions

Let S = ∑ n = 1 ∞ ( − 1 ) n + 1 . 1 n 3 (a) Calculate S n for 1 ≤ n ≤ 10 (b) Use S n − | S n − S n | < b n + 1 to show that

Let $S = \sum_{n = 1}^{\infty} {(- 1)}^{n + 1} . \frac{1}{n^{3}}$

(a) Calculate $S_{n}$ for $1 \leq n \leq 10$

(b) Use $S_{n} - | S_{n} - S_{n} | < b_{n + 1}$ to show that