Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 9: Problem 17
Find the radius of convergence. $$\sum_{n=1}^{\infty} \frac{(x-3)^{n}}{n 2^{n}}$$
Short Answer
Expert verified
The radius of convergence is 2.
Step by step solution
01
Identify the general term
Observe the given series \( \sum_{n=1}^{\infty} \frac{(x-3)^{n}}{n 2^{n}} \). Identify its general term as \( a_n = \frac{(x-3)^n}{n 2^n} \).
02
Setting up the Ratio Test
To find the radius of convergence, use the Ratio Test. Compute the limit \( \lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| \). For our series, this becomes \( \frac{|(x-3)^{n+1}|}{(n+1)2^{n+1}} \div \frac{|(x-3)^{n}|}{n 2^{n}} \).
03
Simplify the ratio
Simplify \( \left| \frac{a_{n+1}}{a_n} \right| \):\[ \left| \frac{(x-3)^{n+1}}{(n+1)2^{n+1}} \times \frac{n 2^n}{(x-3)^n} \right| = \frac{|x-3|}{2} \times \frac{n}{n+1}. \]
04
Take the limit
Now, find the limit as \( n \to \infty \):\[ \lim_{{n \to \infty}} \left( \frac{|x-3|}{2} \times \frac{n}{n+1} \right) = \frac{|x-3|}{2} \cdot \lim_{{n \to \infty}} \frac{n}{n+1} = \frac{|x-3|}{2}. \]
05
Set the limit less than 1
For the Ratio Test to indicate convergence, set \( \frac{|x-3|}{2} < 1 \). This gives us \[ |x-3| < 2. \]
06
Solve for the radius
The inequality \( |x-3| < 2 \) indicates that the radius of convergence is 2.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Ratio Test
The Ratio Test is a handy tool in determining the radius of convergence for a power series. It helps us analyze whether a series converges or diverges by considering the limit of the ratio of consecutive terms. To apply the Ratio Test, we start with a series
- Let the general term be denoted as \( a_n \).
- Compute the absolute value of the ratio \( \left| \frac{a_{n+1}}{a_n} \right| \).
- Take the limit as \( n \to \infty \).
- If the limit is less than 1, the series converges.
Power Series
A power series is a series of the form \( \sum_{n=0}^{\infty} a_n (x - c)^n \), where \( a_n \) are coefficients and \( c \) is the center of the series. It resembles an infinite polynomial where the powers of \((x - c)\) control the terms. The concept is critically important because it helps describe functions and analyze their behavior around different points.
- The expression in our example, \( \frac{(x-3)^n}{n 2^n} \), can be seen as a power series centered at \( x = 3 \).
- The convergence of a power series depends heavily on the distance \( |x - c| \), which dictates how far \( x \) can vary from the center while maintaining convergence.
Limit of a Sequence
The concept of a limit is crucial when analyzing the convergence of series, especially in the Ratio Test. A limit of a sequence is a value that the sequence approaches as the index \( n \) becomes infinitely large. It allows us to understand long-term behavior without calculating all terms individually.Finding limits involves specific steps:
- Identify the sequence \( b_n \).
- Find \( \lim_{{n \to \infty}} b_n \).
- Simplify, if necessary, and determine its behavior.
- If the sequence approaches a finite number, the limit exists; if not, it may diverge.
Convergence Criteria
Convergence criteria help us determine if a series sums to a finite value or diverges to infinity. For the Ratio Test, a series converges if the limit we find is strictly less than 1. Once the limit of the ratio is computed, the convergence criteria transform into an inequality.In our exercise example:
- The expression simplifies to \( \frac{|x-3|}{2} \).
- The criteria are set: enforce \( \frac{|x-3|}{2} < 1 \).
- By solving \( |x-3| < 2 \), derive the interval where the series converges.
- The radius of convergence is the length of this interval, meaning the distance from the center where the series maintains convergence.
