Problem 38 - Thomas Calculus

Question

In Exercises \(37-40,\) find the series' radius of convergence. $$ \sum_{n=1}^{\infty}\left(\frac{2 \cdot 4 \cdot 6 \cdots(2 n)}{2 \cdot 5 \cdot 8 \cdots(3 n-1)}\right)^{2} x^{n} $$

Step-by-Step Solution

Answer

The radius of convergence is found using the ratio test.

Step 1: Identify the General Term

First, we identify the general term of the series: this is given by \( \left( \frac{2 \cdot 4 \cdot 6 \cdots(2n)}{2 \cdot 5 \cdot 8 \cdots(3n-1)} \right)^2 x^n \). The top is the product of even numbers up to \(2n\), and the bottom is the sequence that increases by 3 starting from 2.

Step 2: Express with Factorials

Convert the products into factorial expressions. Notice that the numerator is \((2n)!!\), the product of the first even numbers, and the denominator is more complex but similar to a factorial. Write it out explicitly: \((2n)! / (n!)^2\) and \((3n-1)!!\).

Step 3: Simplify the General Term

Using the factorial expressions, simplify the term to make it easier to analyze. The original fraction becomes \(\frac{(2n)!}{((3n-1)!!)^2} \), and thus the general term of the sequence is \(\left(\frac{(2n)!}{((3n-1)!!)^2}\right)^2 x^n\).

Step 4: Apply the Ratio Test

To find the radius of convergence, apply the ratio test, which involves the limit \( \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| \). Substitute in the general term for \(a_n\). The goal with the ratio test is to cancel out terms as much as possible.

Step 5: Simplify Ratio and Calculate the Limit

Simplifying the ratio gives us \(\left|\frac{(2n+2)! x}{(3n+2)^2 ((3n-1)(3n-2))}\right|\), which simplifies further. Simplify and find the limit as \(n \to \infty\). This will involve using Stirling's approximation or recognizing growth rates.

Step 6: Determination of Radius of Convergence

The series converges if \(|x| < R\) where \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1\) gives \(R\). Typically, equate above limit to \( |x|\) and solve for the condition providing convergence.

Key Concepts

Ratio Test

The Ratio Test is a valuable tool in determining the radius of convergence for a series. It helps us see how the terms in a series behave as they approach infinity. To use the Ratio Test, consider the series \( \sum a_n \), where \( a_n \) is a general term of the series. The test involves calculating the limit \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). If this limit is less than 1, the series absolutely converges. If it's greater than 1, the series diverges. When the limit equals 1, the test is inconclusive.In the context of this exercise, the general term \( a_n \) is a function of factorial expressions, which makes simplification crucial. The Ratio Test requires us to derive \( a_{n+1} \) from \( a_n \) and then simplify the expression \( \frac{a_{n+1}}{a_n} \) before taking the limit. Doing so helps us find the value of \( x \) for which the series converges. This value, \( R \), forms the boundary for convergence and divergence, known as the radius of convergence.

Infinite Series

An infinite series is a sum of an infinite sequence of terms. In mathematical notation, an infinite series is written as \( \sum_{n=1}^{\infty} a_n \), where each \( a_n \) represents the terms to be summed.Such series can behave in varied ways, sometimes converging to a finite value and sometimes diverging. Convergence means adding more terms leads closer to a specific value, whereas divergence means the sum grows without bound as more terms are added.An important aspect of handling infinite series is determining their convergence or divergence. For this task, techniques like the Ratio Test are applied to analyze the infinite behavior of the series, ensuring that \( \sum a_n \) converges within certain boundaries determined by \( |x| < R \), where \( R \) is the radius of convergence.This mathematical analysis finds practical applications across multiple scientific disciplines, proving crucial in fields ranging from engineering to physics, where modelling continuous systems or processes involves infinite sums.

Factorial Sequences

Factorial sequences are often seen in series involving terms with factorial expressions, like \( n! \) or products of sequences similar to factorials, such as double factorials \( (2n)!! \). In our exercise, the sequence features both, forming a complex term.Factorials grow very rapidly compared to regular products, as they multiply all integer values up to a number \( n \). Understanding their behavior helps simplify expressions in series and is essential for applying tests like the Ratio Test. Factorial sequences are written using exclamation marks for ease: \( n! = n \times (n-1) \times \ldots \times 1 \).In more complex sequences, pieces like double factorials account for even or odd number products alone, such as \( (2n)!! = 2 \times 4 \times 6 \times \cdots \times 2n \). Knowing how to simplify these using basic properties like \( (2n)! \approx \sqrt{2\pi n} \left( \frac{n}{e} \right)^n \) via Stirling’s approximation is vital in evaluating the behavior of sequences and determining convergence through the Ratio Test.