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How did you get the (2n+1)(2n-1)? I just don't know where the (2n+1) came from. thanks in advance

How did you get the (2n+1)(2n-1)? I just don't know where the (2n+1) came from. thanks in advance

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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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.
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.
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