Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Question
**Problem Statement:** Find the 5th partial sum, \( s_5 \), of the series \[ \sum_{n=2}^{\infty} (-1)^n \] **Solution:** \[ s_5 = \boxed{0} \] **Feedback:** The answer provided, \( s_5 = 0 \), is incorrect (indicated by a red cross). To find the correct 5th partial sum of the series \(\sum_{n=2}^{n=6} (-1)^n\), sum the terms from \(n = 2\) to \(n = 6\): 1. \(n = 2 \Rightarrow (-1)^2 = 1\) 2. \(n = 3 \Rightarrow (-1)^3 = -1\) 3. \(n = 4 \Rightarrow (-1)^4 = 1\) 4. \(n = 5 \Rightarrow (-1)^5 = -1\) 5. \(n = 6 \Rightarrow (-1)^6 = 1\) Adding these values gives: \[ 1 + (-1) + 1 + (-1) + 1 = 1 \] Therefore, the correct value for \( s_5 \) should be \( \boxed{1} \).
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Transcribed Image Text:**Problem Statement:** Find the 5th partial sum, \( s_5 \), of the series \[ \sum_{n=2}^{\infty} (-1)^n \] **Solution:** \[ s_5 = \boxed{0} \] **Feedback:** The answer provided, \( s_5 = 0 \), is incorrect (indicated by a red cross). To find the correct 5th partial sum of the series \(\sum_{n=2}^{n=6} (-1)^n\), sum the terms from \(n = 2\) to \(n = 6\): 1. \(n = 2 \Rightarrow (-1)^2 = 1\) 2. \(n = 3 \Rightarrow (-1)^3 = -1\) 3. \(n = 4 \Rightarrow (-1)^4 = 1\) 4. \(n = 5 \Rightarrow (-1)^5 = -1\) 5. \(n = 6 \Rightarrow (-1)^6 = 1\) Adding these values gives: \[ 1 + (-1) + 1 + (-1) + 1 = 1 \] Therefore, the correct value for \( s_5 \) should be \( \boxed{1} \).
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