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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![**Analyzing Convergence of a Sequence**
**Problem Statement:**
Does the sequence converge? If so, what is its limit? If not, why not?
\[ \left\{ \frac{ne^{-n}}{n + e^{-n}} \right\}_{n=1}^{\infty} \]
**Explanation:**
We are given a sequence defined by the formula \( \frac{ne^{-n}}{n + e^{-n}} \) and we need to determine whether this sequence converges as \( n \) approaches infinity. Convergence of a sequence means that as \( n \) becomes very large, the sequence approaches some finite limit.
To analyze this sequence, observe the following:
1. **Numerator Analysis**:
- The term \( ne^{-n} \) in the numerator can be broken down as \( n \times e^{-n} \).
- As \( n \to \infty \), \( e^{-n} \) (which is equivalent to \( \frac{1}{e^n} \)) decreases rapidly towards 0.
- Meanwhile, \( n \) increases without bound.
2. **Denominator Analysis**:
- The term \( n + e^{-n} \) in the denominator consists of two parts: \( n \) and \( e^{-n} \).
- As \( n \to \infty \), the term \( e^{-n} \) becomes negligible in comparison to \( n \).
3. **Behavior of Fraction**:
- Since \( e^{-n} \to 0 \), both the numerator \( ne^{-n} \) and the additional term in the denominator \( e^{-n} \) in the given formula also approach 0.
- Consequently, the numerator \( ne^{-n} \) decreases faster to 0 than the denominator due to its exponential decline.
Considering these observations, the fraction \( \frac{ne^{-n}}{n + e^{-n}} \) simplifies for large \( n \) as follows:
\[ \frac{ne^{-n}}{n + e^{-n}} \approx \frac{0}{n} = 0 \]
Therefore, as \( n \) approaches infinity, the sequence \( \left\{ \frac{ne^{-n}}{n + e^{-n}} \right\}_{n=1}^{\infty} \) converges](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1300845a-828a-484d-8f06-319293900125%2Fb5bce4ef-5333-4864-8e88-45c4e766d197%2Fvgqs9r_processed.png&w=3840&q=75)
Transcribed Image Text:**Analyzing Convergence of a Sequence**
**Problem Statement:**
Does the sequence converge? If so, what is its limit? If not, why not?
\[ \left\{ \frac{ne^{-n}}{n + e^{-n}} \right\}_{n=1}^{\infty} \]
**Explanation:**
We are given a sequence defined by the formula \( \frac{ne^{-n}}{n + e^{-n}} \) and we need to determine whether this sequence converges as \( n \) approaches infinity. Convergence of a sequence means that as \( n \) becomes very large, the sequence approaches some finite limit.
To analyze this sequence, observe the following:
1. **Numerator Analysis**:
- The term \( ne^{-n} \) in the numerator can be broken down as \( n \times e^{-n} \).
- As \( n \to \infty \), \( e^{-n} \) (which is equivalent to \( \frac{1}{e^n} \)) decreases rapidly towards 0.
- Meanwhile, \( n \) increases without bound.
2. **Denominator Analysis**:
- The term \( n + e^{-n} \) in the denominator consists of two parts: \( n \) and \( e^{-n} \).
- As \( n \to \infty \), the term \( e^{-n} \) becomes negligible in comparison to \( n \).
3. **Behavior of Fraction**:
- Since \( e^{-n} \to 0 \), both the numerator \( ne^{-n} \) and the additional term in the denominator \( e^{-n} \) in the given formula also approach 0.
- Consequently, the numerator \( ne^{-n} \) decreases faster to 0 than the denominator due to its exponential decline.
Considering these observations, the fraction \( \frac{ne^{-n}}{n + e^{-n}} \) simplifies for large \( n \) as follows:
\[ \frac{ne^{-n}}{n + e^{-n}} \approx \frac{0}{n} = 0 \]
Therefore, as \( n \) approaches infinity, the sequence \( \left\{ \frac{ne^{-n}}{n + e^{-n}} \right\}_{n=1}^{\infty} \) converges
![**Does the series converge? If so, what does it converge to? If not, why not?**
\[ \sum_{n=0}^{\infty} \frac{2}{(-5)^n} \]
To determine whether the series converges, check if the series forms a geometric series and evaluate its common ratio. If the common ratio is between -1 and 1 (exclusive), the series converges.
### Analysis:
We can rewrite the given series as:
\[ \sum_{n=0}^{\infty} \frac{2}{(-5)^n} \]
This series is in the form of a geometric series where \( a = 2 \) (the first term) and \( r = \frac{1}{-5} = -\frac{1}{5} \) (the common ratio).
### Convergence Test for Geometric Series:
A geometric series converges if \( |r| < 1 \). For our series:
\[ |r| = \left| -\frac{1}{5} \right| = \frac{1}{5} < 1\]
Since the common ratio \( \left| -\frac{1}{5} \right| \) is less than 1, the series converges.
### Sum of the Convergent Series:
The sum, \( S \), of an infinite geometric series can be found using the formula:
\[ S = \frac{a}{1 - r} \]
Substitute \( a = 2 \) and \( r = -\frac{1}{5} \):
\[ S = \frac{2}{1 - \left( -\frac{1}{5} \right)} = \frac{2}{1 + \frac{1}{5}} = \frac{2}{\frac{6}{5}} = 2 \cdot \frac{5}{6} = \frac{10}{6} = \frac{5}{3} \]
### Conclusion:
The series converges, and its sum is \(\frac{5}{3}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1300845a-828a-484d-8f06-319293900125%2Fb5bce4ef-5333-4864-8e88-45c4e766d197%2Fugjd3yq_processed.png&w=3840&q=75)
Transcribed Image Text:**Does the series converge? If so, what does it converge to? If not, why not?**
\[ \sum_{n=0}^{\infty} \frac{2}{(-5)^n} \]
To determine whether the series converges, check if the series forms a geometric series and evaluate its common ratio. If the common ratio is between -1 and 1 (exclusive), the series converges.
### Analysis:
We can rewrite the given series as:
\[ \sum_{n=0}^{\infty} \frac{2}{(-5)^n} \]
This series is in the form of a geometric series where \( a = 2 \) (the first term) and \( r = \frac{1}{-5} = -\frac{1}{5} \) (the common ratio).
### Convergence Test for Geometric Series:
A geometric series converges if \( |r| < 1 \). For our series:
\[ |r| = \left| -\frac{1}{5} \right| = \frac{1}{5} < 1\]
Since the common ratio \( \left| -\frac{1}{5} \right| \) is less than 1, the series converges.
### Sum of the Convergent Series:
The sum, \( S \), of an infinite geometric series can be found using the formula:
\[ S = \frac{a}{1 - r} \]
Substitute \( a = 2 \) and \( r = -\frac{1}{5} \):
\[ S = \frac{2}{1 - \left( -\frac{1}{5} \right)} = \frac{2}{1 + \frac{1}{5}} = \frac{2}{\frac{6}{5}} = 2 \cdot \frac{5}{6} = \frac{10}{6} = \frac{5}{3} \]
### Conclusion:
The series converges, and its sum is \(\frac{5}{3}\).
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