**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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Description: **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,

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Calculus: Early Transcendentals

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Chapter1: Functions And Models

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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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**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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