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
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
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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![### 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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F01ab8b4d-9f7b-47d6-9013-d88fec313f5d%2F2e7238a2-aa84-409c-b33e-e6a55af0e305%2Ff9sznha_processed.png&w=3840&q=75)
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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