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
![**Problem Statement:**
(1) Find the summation of the series
\[
\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots
\]
**Explanation:**
This is an infinite geometric series where the first term \(a\) is \(\frac{1}{3}\) and the common ratio \(r\) is also \(\frac{1}{3}\).
The sum \(S\) of an infinite geometric series can be found using the formula:
\[
S = \frac{a}{1 - r}
\]
where \(|r| < 1\).
Applying the values from the series:
- \(a = \frac{1}{3}\)
- \(r = \frac{1}{3}\)
Thus,
\[
S = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}
\]
Therefore, the summation of the series is \(\frac{1}{2}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbe73d2f7-f48d-4799-ad26-f906d4f73fa5%2F76396e3f-776c-439a-8753-215ad3975208%2Fmmq094q_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
(1) Find the summation of the series
\[
\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots
\]
**Explanation:**
This is an infinite geometric series where the first term \(a\) is \(\frac{1}{3}\) and the common ratio \(r\) is also \(\frac{1}{3}\).
The sum \(S\) of an infinite geometric series can be found using the formula:
\[
S = \frac{a}{1 - r}
\]
where \(|r| < 1\).
Applying the values from the series:
- \(a = \frac{1}{3}\)
- \(r = \frac{1}{3}\)
Thus,
\[
S = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}
\]
Therefore, the summation of the series is \(\frac{1}{2}\).
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