Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 9-28, find the convergence set for the given power series. Hint: First find a formula for the nth term; then use the Absolute Ratio Test. $$ 1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}-\frac{x^{6}}{6 !}+\frac{x^{8}}{8 !}-\frac{x^{10}}{10 !}+\cdots $$

Step-by-Step Solution

Answer

The convergence set is \( \mathbb{R} \), all real numbers.

Step 1: Identify the General Term

The power series given is: \( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \ldots \).Notice the pattern in this series: the terms alternate in sign and the power of \( x \) is an even number, which increases by 2 for each consecutive term.The general term can be expressed as \( a_n = (-1)^n \frac{x^{2n}}{(2n)!} \).

Step 2: Apply the Absolute Ratio Test

The Absolute Ratio Test states that a series \( \sum a_n \) converges if \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \].Calculate \( \frac{a_{n+1}}{a_n} \) for our series:\( \frac{a_{n+1}}{a_n} = \frac{(-1)^{n+1} \frac{x^{2(n+1)}}{(2(n+1))!}}{(-1)^n \frac{x^{2n}}{(2n)!}} = \left| \frac{x^2}{(2n+2)(2n+1)} \right| \).Simplify to get \( \left| \frac{x^2}{(2n+2)(2n+1)} \right| \).

Step 3: Find the Limit as n Approaches Infinity

We need to find the limit of the expression we obtained in Step 2 as \( n \to \infty \):\[ \lim_{n \to \infty} \left| \frac{x^2}{(2n+2)(2n+1)} \right| = \lim_{n \to \infty} \frac{|x|^2}{(2n+2)(2n+1)} = 0 \].Since this limit is 0 for all \( x \), it is less than 1.

Step 4: Determine the Convergence Set

Given that \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 0 < 1 \) for all \( x \), the series converges for any value of \( x \).Hence, the convergence set is the entire set of real numbers, \( \mathbb{R} \).

Key Concepts

Absolute Ratio Test

The Absolute Ratio Test is a key tool for determining the convergence or divergence of infinite series, especially those within the realm of power series. This test involves examining the limit of the absolute value of the ratio of consecutive terms of the series. To apply the Absolute Ratio Test, you first need to find the expression for the ratio \( \frac{a_{n+1}}{a_n} \). This is done by taking the \( n+1 \)th term of the power series and dividing it by the \( n \)th term. Once the expression is simplified, the next step is to assess its limit as \( n \) approaches infinity. If this limit is less than 1, i.e., \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \), the series is convergent. However, if it is greater than or equal to 1, the series does not converge. This test works beautifully for our series because the limit converges to 0, which is certainly less than 1. Therefore, the series converges for all values of \( x \).

General Term of a Series

Understanding the general term of a series is crucial, as it unravels the pattern of the series and aids in applying convergence tests accurately. The general term is a formula that represents any term in the series based on its position, \( n \), within the sequence. In the case of the series we're examining, its terms follow a distinct pattern with alternating signs and even powers of \( x \), formulated as \( a_n = (-1)^n \frac{x^{2n}}{(2n)!} \). Here:

• \((-1)^n\) indicates the alternating signs of the terms.
• \(x^{2n}\) indicates that the powers of \( x \) are even, increasing by 2.
• The denominator \((2n)!\) consists of the factorial of even numbers, increasing correspondingly with \( n \).

This structured formulation allows us to proceed with convergence testing, precisely using the Absolute Ratio Test, since we now have an expression for any term in the series.

Convergence Set

The convergence set of a power series refers to all the values of \( x \) for which the series converges. Determining the convergence set is crucial, as it reveals the "domain" where the series is well-behaved and yields a finite sum. For some power series, you might find their convergence set to be a limited range of \( x \) values. However, in our exercise, due to the result of the Absolute Ratio Test, the series converges for any real number \( x \). The significant conclusion here is that when \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 0 \), it indicates convergence for all \( x \). Thus, the convergence set is the entire set of real numbers, denoted by \( \mathbb{R} \), highlighting that there aren’t any restrictions on \( x \) in terms of ensuring this series converges.