Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Selecting and Summing Convergent Geometric Series**
The goal is to identify the convergent geometric series from the following options and then determine its sum:
1. \( \sum_{n=1}^{\infty} \frac{1}{n} \)
2. \( \sum_{n=1}^{\infty} 2(1.01)^n \)
3. \( \sum_{n=1}^{\infty} 2(0.99)^{n-1} \)
**Explanation:**
- The first series is the harmonic series, which is known to diverge.
- The second series has a common ratio of 1.01, which is greater than 1, making it a divergent geometric series.
- The third series has a common ratio of 0.99, which is less than 1, indicating that it is a convergent geometric series.
To calculate the sum of the convergent geometric series \( \sum_{n=1}^{\infty} 2(0.99)^{n-1} \), use the formula for the sum of an infinite geometric series:
\[ S = \frac{a}{1 - r} \]
where \( a = 2 \) is the first term and \( r = 0.99 \) is the common ratio.
Thus, the sum is:
\[ S = \frac{2}{1 - 0.99} = \frac{2}{0.01} = 200 \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F85bb737b-6258-4d50-8ea7-e8c7d70d10bd%2F52703c8b-b62c-4533-af64-2490c03f0681%2Feyw0og_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Selecting and Summing Convergent Geometric Series**
The goal is to identify the convergent geometric series from the following options and then determine its sum:
1. \( \sum_{n=1}^{\infty} \frac{1}{n} \)
2. \( \sum_{n=1}^{\infty} 2(1.01)^n \)
3. \( \sum_{n=1}^{\infty} 2(0.99)^{n-1} \)
**Explanation:**
- The first series is the harmonic series, which is known to diverge.
- The second series has a common ratio of 1.01, which is greater than 1, making it a divergent geometric series.
- The third series has a common ratio of 0.99, which is less than 1, indicating that it is a convergent geometric series.
To calculate the sum of the convergent geometric series \( \sum_{n=1}^{\infty} 2(0.99)^{n-1} \), use the formula for the sum of an infinite geometric series:
\[ S = \frac{a}{1 - r} \]
where \( a = 2 \) is the first term and \( r = 0.99 \) is the common ratio.
Thus, the sum is:
\[ S = \frac{2}{1 - 0.99} = \frac{2}{0.01} = 200 \]
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