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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![### Tutorial Exercise
**Exercise: Finding Terms and Summing an Infinite Series**
---
If the \( n \)th partial sum of a series \( \sum_{n=1}^{\infty} a_n \) is
\[ s_n = \frac{n - 19}{n + 19} \]
we aim to find \( a_n \) and
\[ \sum_{n=1}^{\infty} a_n \).
#### Step 1
If the \( n \)th partial sum for \( \sum_{n=1}^{\infty} a_n \) is
\[ s_n = \frac{n - 19}{n + 19} ,\]
then
\[ a_1 = s_1 = \frac{-9}{10} .\]
The calculation is verified to be
\[ \frac{-9}{10} .\]
#### Step 2
In general, we have
\[ a_n = s_n - s_{n-1} \]
for \( n \ne 1 \).
#### Step 3
So,
\[ a_n = \frac{n - 19}{n + 19} - \frac{(n - 1) - 19}{(n - 1) + 19} = \frac{n - 19}{n + 19} - \boxed{\phantom{}} = \boxed{\phantom{}}. \]
\[ \quad \]
\[ \frac{(n + 18)(n + 19)} .\]
---
**Explanation of Terms and Steps:**
1. The given partial sum \( s_n \) describes the sum of the first \( n \) terms of a series.
2. By setting \( n = 1 \), we evaluate \( s_1 \) to find the first term \( a_1 \).
3. For subsequent terms \( a_n \), we use the relationship \( a_n = s_n - s_{n-1} \) to find the general formula.
This exercise involves algebraic manipulation and understanding the properties of series and sequences to find individual terms and the overall sum of an infinite series. The steps illustrated help in breaking down the process for easier understanding.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa516072c-e3e7-4e03-a334-22985e001f4f%2Fd4730acc-3753-479d-b60d-06602144e1b6%2Fsl22vf4.png&w=3840&q=75)
Transcribed Image Text:### Tutorial Exercise
**Exercise: Finding Terms and Summing an Infinite Series**
---
If the \( n \)th partial sum of a series \( \sum_{n=1}^{\infty} a_n \) is
\[ s_n = \frac{n - 19}{n + 19} \]
we aim to find \( a_n \) and
\[ \sum_{n=1}^{\infty} a_n \).
#### Step 1
If the \( n \)th partial sum for \( \sum_{n=1}^{\infty} a_n \) is
\[ s_n = \frac{n - 19}{n + 19} ,\]
then
\[ a_1 = s_1 = \frac{-9}{10} .\]
The calculation is verified to be
\[ \frac{-9}{10} .\]
#### Step 2
In general, we have
\[ a_n = s_n - s_{n-1} \]
for \( n \ne 1 \).
#### Step 3
So,
\[ a_n = \frac{n - 19}{n + 19} - \frac{(n - 1) - 19}{(n - 1) + 19} = \frac{n - 19}{n + 19} - \boxed{\phantom{}} = \boxed{\phantom{}}. \]
\[ \quad \]
\[ \frac{(n + 18)(n + 19)} .\]
---
**Explanation of Terms and Steps:**
1. The given partial sum \( s_n \) describes the sum of the first \( n \) terms of a series.
2. By setting \( n = 1 \), we evaluate \( s_1 \) to find the first term \( a_1 \).
3. For subsequent terms \( a_n \), we use the relationship \( a_n = s_n - s_{n-1} \) to find the general formula.
This exercise involves algebraic manipulation and understanding the properties of series and sequences to find individual terms and the overall sum of an infinite series. The steps illustrated help in breaking down the process for easier understanding.
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