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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![### Finding the Radius and Interval of Convergence for a Series
The inverse hyperbolic tangent function is written as \(\tanh^{-1}(x)\) and has the following Maclaurin series:
\[
\tanh^{-1}(x) = \sum_{n=1}^{\infty} \frac{1}{2n-1} x^{2n-1}
\]
#### Objective:
Find the radius and interval of convergence for this series. Show all work to justify your conclusion.
#### Step-by-Step Solution:
1. **Identify the series:**
\[
\sum_{n=1}^{\infty} \frac{1}{2n-1} x^{2n-1}
\]
2. **Use the Ratio Test for series convergence:**
The Ratio Test states that for a series \(\sum c_n\), the series converges if:
\[
\lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right| < 1
\]
3. **Identify \(c_n\):**
\[
c_n = \frac{1}{2n-1} x^{2n-1}
\]
4. **Compute \(\frac{c_{n+1}}{c_n}\):**
\[
\frac{c_{n+1}}{c_n} = \frac{\frac{1}{2(n+1)-1} x^{2(n+1)-1}}{\frac{1}{2n-1} x^{2n-1}} = \frac{\frac{1}{2n+1} x^{2n+1}}{\frac{1}{2n-1} x^{2n-1}} = \frac{(2n-1) x^{2n+1}}{(2n+1) x^{2n-1}}
\]
Simplifying this, we get:
\[
\frac{(2n-1) x^{2n+1}}{(2n+1) x^{2n-1}} = \frac{(2n-1) x^2}{(2n+1)}
\]
5. **Take the limit \(n \to \infty\):**
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Transcribed Image Text:### Finding the Radius and Interval of Convergence for a Series
The inverse hyperbolic tangent function is written as \(\tanh^{-1}(x)\) and has the following Maclaurin series:
\[
\tanh^{-1}(x) = \sum_{n=1}^{\infty} \frac{1}{2n-1} x^{2n-1}
\]
#### Objective:
Find the radius and interval of convergence for this series. Show all work to justify your conclusion.
#### Step-by-Step Solution:
1. **Identify the series:**
\[
\sum_{n=1}^{\infty} \frac{1}{2n-1} x^{2n-1}
\]
2. **Use the Ratio Test for series convergence:**
The Ratio Test states that for a series \(\sum c_n\), the series converges if:
\[
\lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right| < 1
\]
3. **Identify \(c_n\):**
\[
c_n = \frac{1}{2n-1} x^{2n-1}
\]
4. **Compute \(\frac{c_{n+1}}{c_n}\):**
\[
\frac{c_{n+1}}{c_n} = \frac{\frac{1}{2(n+1)-1} x^{2(n+1)-1}}{\frac{1}{2n-1} x^{2n-1}} = \frac{\frac{1}{2n+1} x^{2n+1}}{\frac{1}{2n-1} x^{2n-1}} = \frac{(2n-1) x^{2n+1}}{(2n+1) x^{2n-1}}
\]
Simplifying this, we get:
\[
\frac{(2n-1) x^{2n+1}}{(2n+1) x^{2n-1}} = \frac{(2n-1) x^2}{(2n+1)}
\]
5. **Take the limit \(n \to \infty\):**
\
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