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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Question
![**Determine the sum of the following series:**
\[ \sum_{n=1}^{\infty} \frac{(4)^{n-1}}{6^n} \]
This is an infinite series where each term is given by the formula \(\frac{(4)^{n-1}}{6^n}\), with \(n\) starting from 1 and going to infinity. The expression involves exponential terms in both the numerator and the denominator.
### Explanation:
The series can be thought of as a geometric series. A geometric series is a series of the form \(\sum_{n=0}^{\infty} ar^n\), where:
- \(a\) is the first term.
- \(r\) is the common ratio between the terms, such that \(0 < |r| < 1\).
In this case, the series starts at \(n=1\), so we can adjust the formula to fit the standard format. Here, the first term \(a\) is \((4)^0 / 6^1\), and the common ratio \(r\) is \(\frac{4}{6} = \frac{2}{3}\).
### Steps to Find the Sum:
1. **Identify the First Term (\(a\)):**
\[
a = \frac{(4)^0}{6^1} = \frac{1}{6}
\]
2. **Identify the Common Ratio (\(r\)):**
\[
r = \frac{4}{6} = \frac{2}{3}
\]
3. **Use the Sum Formula for an Infinite Geometric Series:**
If \(|r| < 1\), the sum \(S\) of the series is given by:
\[
S = \frac{a}{1 - r}
\]
4. **Calculate the Sum:**
\[
S = \frac{\frac{1}{6}}{1 - \frac{2}{3}} = \frac{\frac{1}{6}}{\frac{1}{3}} = \frac{1}{6} \times \frac{3}{1} = \frac{1}{2}
\]
Thus, the sum of the series is \(\frac{1}{2}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F46f560d0-7644-45ad-80cf-bf2f4d1d2273%2F342d19e3-552a-4d44-9848-48cc67dc1484%2Fe1nf9wn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Determine the sum of the following series:**
\[ \sum_{n=1}^{\infty} \frac{(4)^{n-1}}{6^n} \]
This is an infinite series where each term is given by the formula \(\frac{(4)^{n-1}}{6^n}\), with \(n\) starting from 1 and going to infinity. The expression involves exponential terms in both the numerator and the denominator.
### Explanation:
The series can be thought of as a geometric series. A geometric series is a series of the form \(\sum_{n=0}^{\infty} ar^n\), where:
- \(a\) is the first term.
- \(r\) is the common ratio between the terms, such that \(0 < |r| < 1\).
In this case, the series starts at \(n=1\), so we can adjust the formula to fit the standard format. Here, the first term \(a\) is \((4)^0 / 6^1\), and the common ratio \(r\) is \(\frac{4}{6} = \frac{2}{3}\).
### Steps to Find the Sum:
1. **Identify the First Term (\(a\)):**
\[
a = \frac{(4)^0}{6^1} = \frac{1}{6}
\]
2. **Identify the Common Ratio (\(r\)):**
\[
r = \frac{4}{6} = \frac{2}{3}
\]
3. **Use the Sum Formula for an Infinite Geometric Series:**
If \(|r| < 1\), the sum \(S\) of the series is given by:
\[
S = \frac{a}{1 - r}
\]
4. **Calculate the Sum:**
\[
S = \frac{\frac{1}{6}}{1 - \frac{2}{3}} = \frac{\frac{1}{6}}{\frac{1}{3}} = \frac{1}{6} \times \frac{3}{1} = \frac{1}{2}
\]
Thus, the sum of the series is \(\frac{1}{2}\).
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