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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![**Infinite Series Example**
**Instructions:**
Write out the first few terms of the series below. What is the series' sum?
\[ \sum_{n=0}^{\infty} \left( \frac{3}{2^n} + \frac{(-1)^n}{5^n} \right) \]
### Step-by-Step Explanation:
1. **Understanding the Series:**
The given series is an infinite series that starts from \( n=0 \) and continues indefinitely. The series consists of two parts within the summation: \( \frac{3}{2^n} \) and \( \frac{(-1)^n}{5^n} \).
2. **First Few Terms Calculation:**
- For \( n=0 \):
\[ \frac{3}{2^0} + \frac{(-1)^0}{5^0} = 3 + 1 = 4 \]
- For \( n=1 \):
\[ \frac{3}{2^1} + \frac{(-1)^1}{5^1} = \frac{3}{2} - \frac{1}{5} = \frac{15}{10} - \frac{2}{10} = 1.3 \]
- For \( n=2 \):
\[ \frac{3}{2^2} + \frac{(-1)^2}{5^2} = \frac{3}{4} + \frac{1}{25} = \frac{75}{100} + \frac{4}{100} = 0.79 \]
- For \( n=3 \):
\[ \frac{3}{2^3} + \frac{(-1)^3}{5^3} = \frac{3}{8} - \frac{1}{125} = \frac{375 - 8}{1000} = 0.367 \]
Therefore, the first few terms of the series are approximately:
\[ 4, 1.3, 0.79, 0.367, \ldots \]
3. **Finding the Sum:**
The series can be broken into two separate sums:
\[ \sum_{n=0}^{\infty} \frac{3}{2^n} + \sum](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F720535be-d647-4725-8ea5-21875474b49c%2Fbe11ebe5-f219-4433-b4d7-3f1d6f49fedd%2F2oa857n_processed.png&w=3840&q=75)
Transcribed Image Text:**Infinite Series Example**
**Instructions:**
Write out the first few terms of the series below. What is the series' sum?
\[ \sum_{n=0}^{\infty} \left( \frac{3}{2^n} + \frac{(-1)^n}{5^n} \right) \]
### Step-by-Step Explanation:
1. **Understanding the Series:**
The given series is an infinite series that starts from \( n=0 \) and continues indefinitely. The series consists of two parts within the summation: \( \frac{3}{2^n} \) and \( \frac{(-1)^n}{5^n} \).
2. **First Few Terms Calculation:**
- For \( n=0 \):
\[ \frac{3}{2^0} + \frac{(-1)^0}{5^0} = 3 + 1 = 4 \]
- For \( n=1 \):
\[ \frac{3}{2^1} + \frac{(-1)^1}{5^1} = \frac{3}{2} - \frac{1}{5} = \frac{15}{10} - \frac{2}{10} = 1.3 \]
- For \( n=2 \):
\[ \frac{3}{2^2} + \frac{(-1)^2}{5^2} = \frac{3}{4} + \frac{1}{25} = \frac{75}{100} + \frac{4}{100} = 0.79 \]
- For \( n=3 \):
\[ \frac{3}{2^3} + \frac{(-1)^3}{5^3} = \frac{3}{8} - \frac{1}{125} = \frac{375 - 8}{1000} = 0.367 \]
Therefore, the first few terms of the series are approximately:
\[ 4, 1.3, 0.79, 0.367, \ldots \]
3. **Finding the Sum:**
The series can be broken into two separate sums:
\[ \sum_{n=0}^{\infty} \frac{3}{2^n} + \sum
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