Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

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 power series converges for all real numbers, i.e., \((-\infty, \infty)\).

Step 1: Identify the general term

The series given is a power series: \(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots\). We observe that the powers of \(x\) are even and appear as \(2n\), and the factorial in the denominator corresponds to these powers. The general term can be written as \(a_n = \frac{(-1)^n x^{2n}}{(2n)!}\).

Step 2: Apply the Absolute Ratio Test

To apply the Absolute Ratio Test, we consider the ratio \(\left|\frac{a_{n+1}}{a_n}\right|\). We have:\[ a_n = \frac{(-1)^n x^{2n}}{(2n)!}, \quad a_{n+1} = \frac{(-1)^{n+1} x^{2(n+1)}}{(2(n+1))!} \]Calculating the absolute ratio:\[\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{x^{2(n+1)}}{(2(n+1))!} \cdot \frac{(2n)!}{x^{2n}}\right| = \left|\frac{x^2}{(2n+2)(2n+1)}\right|\]As \(n\) approaches infinity, the term \((2n+2)(2n+1)\) becomes very large, making \(\left|\frac{x^2}{(2n+2)(2n+1)}\right|\) approach zero for any real \(x\).

Step 3: Interpret the Ratio Test Result

Since the ratio \(\left|\frac{x^2}{(2n+2)(2n+1)}\right|\) approaches zero for all real \(x\), the series is absolutely convergent for every real \(x\). Therefore, the series converges for all \(x \in \mathbb{R}\).

Step 4: State the convergence set

The convergence set is determined to be all real numbers since the ratio approaches zero for all \(x\). Hence, the series converges for \((-\infty, \infty)\).

Key Concepts

Power Series

A power series is a sum of terms that involve powers of a variable, typically denoted as **x**. Each term in a power series is of the form \(a_n x^n\), where \(a_n\) is a coefficient that may vary with \(n\). In mathematical terms, a power series can be expressed as:

• \(a_0 + a_1x + a_2x^2 + a_3x^3 + \, \ldots \)

In the given exercise, the series is a power series characterized by only even powers of \(x\) in its terms:

• \(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots\).

What is unique about this series is not just the alternating signs but also the factorials in the denominator, which influence the convergence behavior of the series.

Absolute Ratio Test

The Absolute Ratio Test is a fundamental tool for determining the convergence of a series. It allows us to establish whether a series converges absolutely, and is particularly useful for power series. The test utilizes the limit of the absolute value of the ratio of consecutive terms, typically expressed as:

• \[\lim_{{n \to \infty}} \left|\frac{a_{n+1}}{a_n}\right|\]

In our case, the series' terms are given by \(a_n = \frac{(-1)^n x^{2n}}{(2n)!}\). We compute the ratio for consecutive terms to evaluate convergence:

• \[ \left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{x^2}{(2n+2)(2n+1)}\right| \]

As \(n\) approaches infinity, this expression approaches zero, indicating absolute convergence for all real numbers \(x\). Thus, the series converges for every \(x\in \mathbb{R}\).

Factorial in Series

Factorials, denoted by \(n!\), represent the product of all positive integers up to \(n\). In the context of a power series, factorials often appear in the denominator, particularly in series related to exponential functions or trigonometric expansions. Their role is crucial in determining the behavior and convergence of the series, as they grow rapidly with increasing \(n\).
A factorial such as \((2n)!\) in our series contributes significantly to the convergence properties. Since the factorial grows faster than powers of \(n\), they tend to dominate the behavior of the term \(\frac{1}{(2n)!}\), which ensures that each successive term of the series becomes smaller, thus aiding convergence.

General Term of a Series

The general term of a series provides a formulaic representation for each term within the series. It allows us to construct any term based on its position \(n\) within the sequence. Identifying the general term is often the first step in studying the convergence of a series.
For the power series in our exercise, the general term identified is \(a_n = \frac{(-1)^n x^{2n}}{(2n)!}\). Here's how this term is structured:

• The factor \((-1)^n\) creates an alternating sign pattern for each term.
• \(x^{2n}\) ensures that only even powers of \(x\) are considered, corresponding with the factorial \((2n)!\).

Understanding the general term helps when applying convergence tests such as the Absolute Ratio Test, to make quantitative assessments about the series' behavior.