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(2n)! (2n + 2)! 10 an+1 10+1 10 (2n +2)(2n + 1) <1 for n 2 1, so strictly decreasing. an

(2n)! (2n + 2)! 10 an+1 10+1 10 (2n +2)(2n + 1) <1 for n 2 1, so strictly decreasing. an

Calculus: Early Transcendentals
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.
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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