Chapter 9: Problem 21
Find the radius of convergence. $$1+2 x+\frac{4 ! x^{2}}{(2 !)^{2}}+\frac{6 ! x^{3}}{(3 !)^{2}}+\frac{8 ! x^{4}}{(4 !)^{2}}+\frac{10 ! x^{5}}{(5 !)^{2}}+\cdots$$
Short Answer
Expert verified
The radius of convergence is \( \frac{1}{4} \).
Step by step solution
01
Identify the Series Pattern
The given series is: \[1 + 2x + \frac{4! x^2}{(2!)^2} + \frac{6! x^3}{(3!)^2} + \frac{8! x^4}{(4!)^2} + \frac{10! x^5}{(5!)^2} + \cdots\]We need to identify the general term of this series. Observe the pattern: the coefficient of each term appears to be \( \frac{(2n)!}{n!^2} \) for \( n \geq 0 \). This gives the general term as \( a_n = \frac{(2n)! x^n}{(n!)^2} \).
02
Apply the Ratio Test
To find the radius of convergence, we use the Ratio Test:\[\text{Limit} = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\]Plugging in the terms:\[\frac{a_{n+1}}{a_n} = \left| \frac{(2n+2)!}{(n+1)!^2} \cdot \frac{(n!)^2}{(2n)!} x \right|\]Simplify using properties of factorials.
03
Simplify the Ratio
Simplifying the ratio:\[= \left| \frac{(2n+1)(2n+2)}{(n+1)^2} x \right|\]As \( n \to \infty \), divide the terms by \( n^2 \), leading to:\[= \left| \frac{4}{1} x \right| = 4|x|\]
04
Determine Convergence from the Limit
For the series to converge, the result from the Ratio Test must be less than 1:\[4|x| < 1\]Solve this inequality to find:\[|x| < \frac{1}{4}\]Hence, the radius of convergence \( R \) is \( \frac{1}{4} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Ratio Test
The Ratio Test is a powerful tool used to determine the radius of convergence for infinite series. When applied to series of the form \(a_n x^n\), it helps us understand where the series converges or diverges. The Ratio Test examines the limit of the absolute value of the ratio between consecutive terms, specifically \( \lim_{n \to \infty} \left| \frac{a_{n+1} x^{n+1}}{a_n x^n} \right| \). For convergence, we want this limit to be less than 1.
In our example series:
In our example series:
- The general term \( a_n = \frac{(2n)!}{(n!)^2} x^n \).
- By plugging this into the Ratio Test formula, we focus on the part \( \lim_{n \to \infty} \left| \frac{(2n+2)!}{(n+1)!^2} x \cdot \frac{(n!)^2}{(2n)!} \right| \).
- This calculation simplifies down to \(|x \cdot \frac{(2n+1)(2n+2)}{(n+1)^2}| \), which eventually simplifies to \(4|x| \) as \( n \to \infty \).
Factorial
Factorials, denoted by \( n! \), are crucial in combinatorics and series expansions. A factorial is the product of all positive integers up to \( n \). For example, \(4! = 4 \times 3 \times 2 \times 1 = 24\). In the context of the given series, factorials appear extensively in the numerator and denominator of the term coefficients.
For the general term \( a_n = \frac{(2n)! x^n}{(n!)^2} \), the factorials \((2n)!\) and \((n!)^2\) define how terms grow rapidly. We simplify the expressions using factorial properties to ease calculations:
For the general term \( a_n = \frac{(2n)! x^n}{(n!)^2} \), the factorials \((2n)!\) and \((n!)^2\) define how terms grow rapidly. We simplify the expressions using factorial properties to ease calculations:
- The expression \((2n)!\) expands to \((2n)(2n-1)...3 \times 2 \times 1\).
- The \((n!)^2\) term is simply \((n!)\) multiplied by itself.
Series
An infinite series is essentially a sum of terms that can extend indefinitely. The series can be denoted as \( \sum_{n=0}^{\infty} a_n \), where \( a_n \) represents the individual terms. Our aim is to explore whether these sums converge, and if they do, over what interval of \( x \) this convergence holds.
In the original exercise:
In the original exercise:
- The series given is \(1 + 2x + \frac{4! x^2}{(2!)^2} + \cdots\).
- It's derived from a pattern tied to coefficients of \( \frac{(2n)!}{(n!)^2} \).
Convergence
Convergence in mathematics refers to the behavior of a series as it extends toward infinity. Specifically, a series converges if the sums of its terms tend towards a single number. The Radius of Convergence tells us for which values of \( x \) a series converges.
Analyzing convergence involves checking whether the ratio of consecutive terms, as discussed in the Ratio Test section, falls below 1 for values of \( x \). This check helps us determine the series behavior at different points:
Analyzing convergence involves checking whether the ratio of consecutive terms, as discussed in the Ratio Test section, falls below 1 for values of \( x \). This check helps us determine the series behavior at different points:
- In the given example, simplifying the ratio test limit to \(4|x| \) must yield a value less than 1 for convergence.
- This establishes convergence when \(|x| < \frac{1}{4}\), providing us the radius of convergence.