What is the value of p = lim+¹ when applying the Ratio Test to assess n→∞an a convergence of? What is the conclusion of the test? n=134 A. 1; test is inconclusive B. 0; convergent C. 0; divergent D. ~; divergent

Calculus: Early Transcendentals
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Author:James Stewart
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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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### Topic: Convergence of Series Using the Ratio Test

#### Problem Statement

What is the value of 
\[ \rho = \lim_{{n \to \infty}} \left| \frac{{a_{n+1}}}{{a_n}} \right| \]
when applying the Ratio Test to assess the convergence of 
\[ \sum_{{n=1}}^{\infty} \frac{6^n}{3^{n^2}}? \]
What is the conclusion of the test?

#### Multiple Choice Options

A. 1; test is inconclusive

B. 0; convergent

C. 0; divergent

D. ∞; divergent

#### Explanation

The ratio test is a method used to determine the absolute convergence of infinite series. The test involves taking the limit of the absolute value of the ratio of consecutive terms. Specifically, if 
\[ a_n \]
represents the terms of the series, then

\[ \rho = \lim_{{n \to \infty}} \left| \frac{{a_{n+1}}}{{a_n}} \right| \]

Based on the value of 
\[ \rho \],
the series can be classified as follows:
- If 
\[ \rho < 1 \],
the series converges absolutely.
- If 
\[ \rho > 1 \],
the series diverges.
- If 
\[ \rho = 1 \],
the ratio test is inconclusive.

Let's compute the ratio for the given series 
\[ \sum_{{n=1}}^{\infty} \frac{6^n}{3^{n^2}} \].

#### Mathematical Solution
Define 
\[ a_n = \frac{6^n}{3^{n^2}} \].
Then,

\[ \frac{a_{n+1}}{a_n} = \frac{\frac{6^{n+1}}{3^{(n+1)^2}}}{\frac{6^n}{3^{n^2}}} = \frac{6^{n+1}}{3^{(n+1)^2}} \cdot \frac{3^{n^2}}{6^n} = \frac{6 \cdot 6^n}{3^{n^2 + 2n + 1}} \cdot \frac{3^{n^2}}{6
Transcribed Image Text:### Topic: Convergence of Series Using the Ratio Test #### Problem Statement What is the value of \[ \rho = \lim_{{n \to \infty}} \left| \frac{{a_{n+1}}}{{a_n}} \right| \] when applying the Ratio Test to assess the convergence of \[ \sum_{{n=1}}^{\infty} \frac{6^n}{3^{n^2}}? \] What is the conclusion of the test? #### Multiple Choice Options A. 1; test is inconclusive B. 0; convergent C. 0; divergent D. ∞; divergent #### Explanation The ratio test is a method used to determine the absolute convergence of infinite series. The test involves taking the limit of the absolute value of the ratio of consecutive terms. Specifically, if \[ a_n \] represents the terms of the series, then \[ \rho = \lim_{{n \to \infty}} \left| \frac{{a_{n+1}}}{{a_n}} \right| \] Based on the value of \[ \rho \], the series can be classified as follows: - If \[ \rho < 1 \], the series converges absolutely. - If \[ \rho > 1 \], the series diverges. - If \[ \rho = 1 \], the ratio test is inconclusive. Let's compute the ratio for the given series \[ \sum_{{n=1}}^{\infty} \frac{6^n}{3^{n^2}} \]. #### Mathematical Solution Define \[ a_n = \frac{6^n}{3^{n^2}} \]. Then, \[ \frac{a_{n+1}}{a_n} = \frac{\frac{6^{n+1}}{3^{(n+1)^2}}}{\frac{6^n}{3^{n^2}}} = \frac{6^{n+1}}{3^{(n+1)^2}} \cdot \frac{3^{n^2}}{6^n} = \frac{6 \cdot 6^n}{3^{n^2 + 2n + 1}} \cdot \frac{3^{n^2}}{6
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