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Which of the following is an example of a divergent infinite series whose general term tends to zero? 1 Α.Σ. n=1 9 10 ܕ

Which of the following is an example of a divergent infinite series whose general term tends to zero? 1 Α.Σ. n=1 9 10 ܕ

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.3: Geometric Sequences
Problem 81E
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**Title: Understanding Divergent Infinite Series in Calculus** **Content:** **Question:** Which of the following is an example of a divergent infinite series whose general term tends to zero? **Options:** A. \(\sum_{n=1}^{\infty} \frac{1}{n^{\frac{9}{10}}}\) B. \(\sum_{n=1}^{\infty} \frac{n^{\frac{9}{10}}}{1 + n^{\frac{9}{10}}}\) C. \(\sum_{n=1}^{\infty} 2^{-n}\) D. \(\sum_{n=1}^{\infty} n^{\frac{9}{10}}\) **Explanation:** To determine which of these series diverges while its general term tends to zero, we should assess the nature of each series: - **Option A:** \(\sum_{n=1}^{\infty} \frac{1}{n^{\frac{9}{10}}}\) This is a p-series with \(p = \frac{9}{10}\), which is less than 1. According to the p-series test, the series diverges. - **Option B:** \(\sum_{n=1}^{\infty} \frac{n^{\frac{9}{10}}}{1 + n^{\frac{9}{10}}}\) Here, the general term is equivalent to \( \frac{n^{\frac{9}{10}}}{n^{\frac{9}{10}}} = 1 \) for large \( n \), which does not tend to zero. Thus, it cannot be considered for divergence based on the condition given. - **Option C:** \(\sum_{n=1}^{\infty} 2^{-n}\) This is a geometric series with a common ratio of \( \frac{1}{2} \) (which is less than 1), and thus converges. - **Option D:** \(\sum_{n=1}^{\infty} n^{\frac{9}{10}}\) In this case, the general term \( n^{\frac{9}{10}} \) does not tend to zero as \( n \) increases, implying the series diverges but does not meet the requirement that
Transcribed Image Text:**Title: Understanding Divergent Infinite Series in Calculus** **Content:** **Question:** Which of the following is an example of a divergent infinite series whose general term tends to zero? **Options:** A. \(\sum_{n=1}^{\infty} \frac{1}{n^{\frac{9}{10}}}\) B. \(\sum_{n=1}^{\infty} \frac{n^{\frac{9}{10}}}{1 + n^{\frac{9}{10}}}\) C. \(\sum_{n=1}^{\infty} 2^{-n}\) D. \(\sum_{n=1}^{\infty} n^{\frac{9}{10}}\) **Explanation:** To determine which of these series diverges while its general term tends to zero, we should assess the nature of each series: - **Option A:** \(\sum_{n=1}^{\infty} \frac{1}{n^{\frac{9}{10}}}\) This is a p-series with \(p = \frac{9}{10}\), which is less than 1. According to the p-series test, the series diverges. - **Option B:** \(\sum_{n=1}^{\infty} \frac{n^{\frac{9}{10}}}{1 + n^{\frac{9}{10}}}\) Here, the general term is equivalent to \( \frac{n^{\frac{9}{10}}}{n^{\frac{9}{10}}} = 1 \) for large \( n \), which does not tend to zero. Thus, it cannot be considered for divergence based on the condition given. - **Option C:** \(\sum_{n=1}^{\infty} 2^{-n}\) This is a geometric series with a common ratio of \( \frac{1}{2} \) (which is less than 1), and thus converges. - **Option D:** \(\sum_{n=1}^{\infty} n^{\frac{9}{10}}\) In this case, the general term \( n^{\frac{9}{10}} \) does not tend to zero as \( n \) increases, implying the series diverges but does not meet the requirement that
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