Problem 39 - Thomas Calculus

Question

In Exercises \(37-40,\) find the series' radius of convergence. $$ \sum_{n=1}^{\infty} \frac{(n !)^{2}}{2^{n}(2 n) !} x^{n} $$

Step-by-Step Solution

Answer

The radius of convergence of the series is 8.

Step 1: Identify the general term of the series

The general term of the series is given by \(a_n = \frac{(n!)^2}{2^n (2n)!} x^n\). This is the term we will analyze to find the radius of convergence.

Step 2: Apply the Ratio Test

To determine the radius of convergence, apply the Ratio Test: consider \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\). For this series, compute \[\frac{a_{n+1}}{a_n} = \frac{((n+1)!)^2}{2^{n+1} (2(n+1))!} x^{n+1} \cdot \frac{2^n (2n)!}{(n!)^2 x^n} = \frac{((n+1)!)^2}{n!^2 2 (2n+2)(2n+1)} x \cdot \frac{(2n)!}{(2n+2)!}\].

Step 3: Simplify the Ratio

Simplify the ratio obtained:\[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(n+1)^2}{2((2n+2)(2n+1))} x \right|\].The factorials cancel as follows: Normalize the terms in the factorial, and we get the result.

Step 4: Evaluate the Limit

Evaluate the limit as \(n\) approaches infinity:\[\lim_{n \to \infty} \left| \frac{(n+1)^2}{2((2n+2)(2n+1))} \right| = \lim_{n \to \infty} \left| \frac{n^2 + 2n + 1}{8n^2 + 8n} \right| = \frac{1}{8}\].

Step 5: Determine the Radius of Convergence

According to the Ratio Test, the series converges when \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1\). Therefore, set:\[\left| x \right| \cdot \frac{1}{8} < 1 \]Solving for \(x\), we get \(\left| x \right| < 8\). Thus, the radius of convergence \(R\) is \(8\).

Key Concepts

Ratio Test

The Ratio Test is a powerful tool to determine the convergence of an infinite series. When dealing with power series, such as the one presented in our exercise, the Ratio Test helps us evaluate the series' radius of convergence, which indicates where the series converges on the real number line. You start by examining the limit of the absolute value of the ratio of consecutive terms: \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]. Here, \(a_n\) represents the terms of the series. If this limit is less than 1, the series converges absolutely. If the limit is greater than 1, the series diverges. If it equals 1, the test is inconclusive.In our specific problem, we found the limit: \[ \lim_{n \to \infty} \left| \frac{(n+1)^2}{2((2n+2)(2n+1))} \right| = \frac{1}{8} \]. This means that for our series, it converges when the term\(| x | \cdot \frac{1}{8} < 1\). Solving this gives us \(| x | < 8\), revealing a radius of convergence of 8.Understanding how to apply the Ratio Test helps streamline finding convergence intervals in many calculus problems, especially those involving power series.

Power Series

A power series is an infinite series of the form: \[ \sum_{n=0}^{\infty} a_n(x-c)^n \], where \(a_n\) are coefficients, \(x\) is a variable, and \(c\) is the center of the series. One of the essential aspects of power series is determining where they converge, known as the radius of convergence.In practical terms, a power series behaves like a polynomial with infinitely many terms. It is centered at \(c\) and usually converges within some radius on the number line. For instance, in our exercise: \[ \sum_{n=1}^{\infty} \frac{(n!)^2}{2^n(2n)!} x^n \], shows convergence daily by utilizing tests like the Ratio Test.Power series are vital in mathematics as they form the foundation for Taylor series and Fourier series, which are crucial for analyzing functions that might otherwise be difficult to work with. Mastery of power series concepts, including understanding convergence and radius of convergence, is fundamental in advanced calculus and real analysis.

Factorial Simplification

Simplifying factorial expressions can make the process of finding limits or applying tests like the Ratio Test much simpler. Factorials are products of all positive integers up to a specified number. For example, \(n! = n \cdot (n-1) \cdot (n-2) \cdots 1\). In the context of our problem, you often see these occurring in both the numerator and denominator. Simplification becomes key.In the exercise, the general term \(a_n = \frac{(n!)^2}{2^n (2n)!} x^n\) includes factorials, necessitating their simplification. For the Ratio Test applied here, you must find \( \frac{a_{n+1}}{a_n} \), resulting in terms like \(\frac{((n+1)!)^2}{(n!)^2}\) and \(\frac{(2n)!}{(2n+2)!}\). These seem complex, but observing patterns allows simplification:

• \((n+1)! = (n+1)n!\)
• \((2n+2)! = (2n+2)(2n+1)(2n)!\)

These lead to easier manipulation when applying the Ratio Test and produce neat results, such as canceling numerous terms, leading to expressions that simplify the evaluation of limits. Understanding factorial simplifications is essential for success in calculus and helps tackle complex problems involving series.