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:**
15. Find the sum of
\[ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \cdots \]
**Explanation:**
The given mathematical expression is an infinite series. The terms of this series are expressed in the form:
\[ \frac{1}{(2n-1)(2n+1)} \]
where \( n \) is a positive integer starting from 1.
So, the series can be written as:
\[ \sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n+1)} \]
**Steps to Solve:**
1. **Express each term in a partial fraction form:**
- We can rewrite the general term \(\frac{1}{(2n-1)(2n+1)}\) using partial fractions.
\[ \frac{1}{(2n-1)(2n+1)} = \frac{A}{2n-1} + \frac{B}{2n+1} \]
Solving for \( A \) and \( B \):
\[ 1 = A(2n+1) + B(2n-1) \]
Setting \[ n \to \infty \], it can be found that:
\[ A = \frac{1}{2} \] and \[ B = -\frac{1}{2} \]
So:
\[ \frac{1}{(2n-1)(2n+1)} = \frac{1}{2} \left( \frac{1}{2n-1} - \frac{1}{2n+1} \right) \]
2. **Sum of the series:**
- Substituting this back into the series:
\[ \sum_{n=1}^{\infty} \frac{1}{2} \left( \frac{1}{2n-1} - \frac{1}{2n+1} \right) \]
- This is a telescoping series, which means that most of the terms will cancel each other out.
\[ \sum_{n=1}^{\infty} \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F37708386-d0d8-46c3-a2d8-8e337c754037%2F85c1cc6b-8ed6-402c-a90f-611646cac38c%2Fy16usj7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
15. Find the sum of
\[ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \cdots \]
**Explanation:**
The given mathematical expression is an infinite series. The terms of this series are expressed in the form:
\[ \frac{1}{(2n-1)(2n+1)} \]
where \( n \) is a positive integer starting from 1.
So, the series can be written as:
\[ \sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n+1)} \]
**Steps to Solve:**
1. **Express each term in a partial fraction form:**
- We can rewrite the general term \(\frac{1}{(2n-1)(2n+1)}\) using partial fractions.
\[ \frac{1}{(2n-1)(2n+1)} = \frac{A}{2n-1} + \frac{B}{2n+1} \]
Solving for \( A \) and \( B \):
\[ 1 = A(2n+1) + B(2n-1) \]
Setting \[ n \to \infty \], it can be found that:
\[ A = \frac{1}{2} \] and \[ B = -\frac{1}{2} \]
So:
\[ \frac{1}{(2n-1)(2n+1)} = \frac{1}{2} \left( \frac{1}{2n-1} - \frac{1}{2n+1} \right) \]
2. **Sum of the series:**
- Substituting this back into the series:
\[ \sum_{n=1}^{\infty} \frac{1}{2} \left( \frac{1}{2n-1} - \frac{1}{2n+1} \right) \]
- This is a telescoping series, which means that most of the terms will cancel each other out.
\[ \sum_{n=1}^{\infty} \
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