(5x)" . Given the power series Σ(-1)", , find and justify its interval of 4n In(n)' n=2 onvergence and the behavior of the series at the end points of the interval.
(5x)" . Given the power series Σ(-1)", , find and justify its interval of 4n In(n)' n=2 onvergence and the behavior of the series at the end points of the interval.
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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![### Problem 8: Interval of Convergence for a Power Series
#### Given Problem Statement:
Given the power series
\[ \sum_{n=2}^{\infty} \frac{(-1)^n (5x)^n}{4^n \ln(n)} \],
find and justify its interval of convergence and the behavior of the series at the end points of the interval.
#### Steps to Solve the Problem:
1. **Ratio Test for Convergence:**
The ratio test will be used to find the interval of convergence. Given a general term \( a_n = \frac{(-1)^n (5x)^n}{4^n \ln(n)} \), we apply the ratio test:
\[
\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{ \frac{(-1)^{n+1} (5x)^{n+1}}{4^{n+1} \ln(n+1)} }{ \frac{(-1)^n (5x)^n}{4^n \ln(n)} } \right| = \left| \frac{(5x)^{n+1}}{4^{n+1} \ln(n+1)} \cdot \frac{4^n \ln(n)}{(5x)^n} \right|
\]
Simplifying this, we get:
\[
\left| \frac{(5x)(5x)^n}{4 \cdot 4^n \ln(n+1)} \cdot \frac{4^n \ln(n)}{(5x)^n} \right| = \left| \frac{5x}{4} \cdot \frac{\ln(n)}{\ln(n+1)} \right|
\]
As \( n \to \infty \), \(\frac{\ln(n)}{\ln(n+1)} \to 1\). Hence,
\[
\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{5x}{4} \right|
\]
For convergence, the limit should be less than 1:
\[
\left| \frac{5x}{4} \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F197e26df-0092-4237-9a65-d284350ca22d%2Fc7aae577-dad2-4877-8c4c-35bda1dd8036%2Fkddi098_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem 8: Interval of Convergence for a Power Series
#### Given Problem Statement:
Given the power series
\[ \sum_{n=2}^{\infty} \frac{(-1)^n (5x)^n}{4^n \ln(n)} \],
find and justify its interval of convergence and the behavior of the series at the end points of the interval.
#### Steps to Solve the Problem:
1. **Ratio Test for Convergence:**
The ratio test will be used to find the interval of convergence. Given a general term \( a_n = \frac{(-1)^n (5x)^n}{4^n \ln(n)} \), we apply the ratio test:
\[
\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{ \frac{(-1)^{n+1} (5x)^{n+1}}{4^{n+1} \ln(n+1)} }{ \frac{(-1)^n (5x)^n}{4^n \ln(n)} } \right| = \left| \frac{(5x)^{n+1}}{4^{n+1} \ln(n+1)} \cdot \frac{4^n \ln(n)}{(5x)^n} \right|
\]
Simplifying this, we get:
\[
\left| \frac{(5x)(5x)^n}{4 \cdot 4^n \ln(n+1)} \cdot \frac{4^n \ln(n)}{(5x)^n} \right| = \left| \frac{5x}{4} \cdot \frac{\ln(n)}{\ln(n+1)} \right|
\]
As \( n \to \infty \), \(\frac{\ln(n)}{\ln(n+1)} \to 1\). Hence,
\[
\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{5x}{4} \right|
\]
For convergence, the limit should be less than 1:
\[
\left| \frac{5x}{4} \
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