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...
Related questions
Question
How did you get the (2n+1)(2n-1)? I just don't know where the (2n+1) came from.
thanks in advance!
![The text is an analysis of a series to determine its convergence or divergence using the ratio test. Here's the transcription:
---
We have:
\[ u_n = \frac{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1)}{(5^n) \cdot n!} \]
We need to find:
\[ \lim_{n \to \infty} \frac{u_{n+1}}{u_n} = \lim_{n \to \infty} \frac{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1) \cdot (2n+1)}{(5^{n+1}) \cdot (n+1)!} \times \frac{(5^n) \cdot n!}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1)} \]
This simplifies to:
\[ = \lim_{n \to \infty} \frac{(2n+1)}{(5) \cdot (n+1)} \]
\[ = \frac{1}{5} \lim_{n \to \infty} \left(1 + \frac{n}{n+1}\right) \]
\[ = \frac{1}{5} + \lim_{n \to \infty} \frac{1}{1 + \frac{1}{n}} \]
\[ = 1 + \frac{1}{5} > 1 \]
Hence, the series is *divergent*.
---
This explanation systematically breaks down the ratio test calculation, ultimately concluding that the series diverges because the limit is greater than 1.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F42d8c157-6432-40bb-8a52-17ecf5ac12dd%2F84d0b1d3-ddcb-4efe-93cf-a715c496406e%2Fn4jz63_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The text is an analysis of a series to determine its convergence or divergence using the ratio test. Here's the transcription:
---
We have:
\[ u_n = \frac{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1)}{(5^n) \cdot n!} \]
We need to find:
\[ \lim_{n \to \infty} \frac{u_{n+1}}{u_n} = \lim_{n \to \infty} \frac{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1) \cdot (2n+1)}{(5^{n+1}) \cdot (n+1)!} \times \frac{(5^n) \cdot n!}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1)} \]
This simplifies to:
\[ = \lim_{n \to \infty} \frac{(2n+1)}{(5) \cdot (n+1)} \]
\[ = \frac{1}{5} \lim_{n \to \infty} \left(1 + \frac{n}{n+1}\right) \]
\[ = \frac{1}{5} + \lim_{n \to \infty} \frac{1}{1 + \frac{1}{n}} \]
\[ = 1 + \frac{1}{5} > 1 \]
Hence, the series is *divergent*.
---
This explanation systematically breaks down the ratio test calculation, ultimately concluding that the series diverges because the limit is greater than 1.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps with 1 images
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning