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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For this problem, why did we put (2n+2)! In the denominator in the first part , instead of (2n+1)!
![The image displays a mathematical inequality related to a sequence. It is written as follows:
\[
\frac{a_{n+1}}{a_n} = \frac{10^{n+1}}{(2n + 2)!} \cdot \frac{(2n)!}{10^n} = \frac{10}{(2n + 2)(2n + 1)} < 1 \text{ for } n \geq 1, \text{ so strictly decreasing.}
\]
This expression demonstrates the ratio test for sequences, showing that the ratio of consecutive terms is less than 1 for \( n \geq 1 \), indicating that the sequence is strictly decreasing. There are no graphs or diagrams in the image.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F61b75115-d70f-4fe2-af93-2076876ad69a%2F4ae3c99a-e874-45c5-b2ff-30ca5b58566f%2Fj34qi4h_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The image displays a mathematical inequality related to a sequence. It is written as follows:
\[
\frac{a_{n+1}}{a_n} = \frac{10^{n+1}}{(2n + 2)!} \cdot \frac{(2n)!}{10^n} = \frac{10}{(2n + 2)(2n + 1)} < 1 \text{ for } n \geq 1, \text{ so strictly decreasing.}
\]
This expression demonstrates the ratio test for sequences, showing that the ratio of consecutive terms is less than 1 for \( n \geq 1 \), indicating that the sequence is strictly decreasing. There are no graphs or diagrams in the image.
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