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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Contingency Table
A contingency table can be defined as the visual representation of the relationship between two or more categorical variables that can be evaluated and registered. It is a categorical version of the scatterplot, which is used to investigate the linear relationship between two variables. A contingency table is indeed a type of frequency distribution table that displays two variables at the same time.
Binomial Distribution
Binomial is an algebraic expression of the sum or the difference of two terms. Before knowing about binomial distribution, we must know about the binomial theorem.
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![**Understanding the Divergence Test in Series**
The Divergence Test helps us determine whether a given series is divergent. Let us use this tool to analyze the following series:
\[ \sum_{n=1}^\infty \frac{2n+1}{2n^3+8} \]
To apply the Divergence Test, follow these steps:
1. **Examine the terms**: The series given is \(\frac{2n+1}{2n^3+8}\).
2. **Apply the divergence test**: According to the Divergence Test, if the limit of \(\frac{2n+1}{2n^3+8}\) as \(n\) approaches infinity does not equal zero, then the series diverges.
We compute:
\[
\lim_{n \to \infty} \frac{2n+1}{2n^3+8}
\]
As \(n\) approaches infinity, the highest power term in both the numerator and the denominator will dominate, so:
\[
\lim_{n \to \infty} \frac{2n+1}{2n^3+8} = \lim_{n \to \infty} \frac{2n+1}{2n^3} = \lim_{n \to \infty} \frac{2n}{2n^3} + \lim_{n \to \infty} \frac{1}{2n^3} = \lim_{n \to \infty} \frac{2}{2n^2} = \lim_{n \to \infty} \frac{1}{n^2} = 0
\]
Since the limit equals zero, the Divergence Test is inconclusive in proving divergence for this series. Other convergence tests will need to be explored to determine if it converges.
Utilize this approach to check for divergence in other series, and remember, if the limit of the terms does not equal zero, the series must diverge.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7bb6f29c-bae7-4b63-bf8e-21de0c88ea98%2F19530201-2e07-4889-bd9a-3b96bdeef5f6%2Ftv7pdrs.jpeg&w=3840&q=75)
Transcribed Image Text:**Understanding the Divergence Test in Series**
The Divergence Test helps us determine whether a given series is divergent. Let us use this tool to analyze the following series:
\[ \sum_{n=1}^\infty \frac{2n+1}{2n^3+8} \]
To apply the Divergence Test, follow these steps:
1. **Examine the terms**: The series given is \(\frac{2n+1}{2n^3+8}\).
2. **Apply the divergence test**: According to the Divergence Test, if the limit of \(\frac{2n+1}{2n^3+8}\) as \(n\) approaches infinity does not equal zero, then the series diverges.
We compute:
\[
\lim_{n \to \infty} \frac{2n+1}{2n^3+8}
\]
As \(n\) approaches infinity, the highest power term in both the numerator and the denominator will dominate, so:
\[
\lim_{n \to \infty} \frac{2n+1}{2n^3+8} = \lim_{n \to \infty} \frac{2n+1}{2n^3} = \lim_{n \to \infty} \frac{2n}{2n^3} + \lim_{n \to \infty} \frac{1}{2n^3} = \lim_{n \to \infty} \frac{2}{2n^2} = \lim_{n \to \infty} \frac{1}{n^2} = 0
\]
Since the limit equals zero, the Divergence Test is inconclusive in proving divergence for this series. Other convergence tests will need to be explored to determine if it converges.
Utilize this approach to check for divergence in other series, and remember, if the limit of the terms does not equal zero, the series must diverge.
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