Problem 9 - 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. $$ \frac{x}{1 \cdot 2}-\frac{x^{2}}{2 \cdot 3}+\frac{x^{3}}{3 \cdot 4}-\frac{x^{4}}{4 \cdot 5}+\frac{x^{5}}{5 \cdot 6}-\cdots $$

Step-by-Step Solution

Answer

The convergence set is all real numbers.

Step 1: Identify the General Term

The given series is \( \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n(n+1)} \). Each term in the series alternates in sign and has the form \( \frac{(-1)^{n+1} x^n}{n(n+1)} \).

Step 2: Apply the Absolute Ratio Test

To determine the convergence set, we use the Absolute Ratio Test. For the series \( \sum a_n \), the ratio of consecutive terms is given by \( \left| \frac{a_{n+1}}{a_n} \right| \). Compute the absolute value for \( a_n = \frac{x^n}{n(n+1)} \) and \( a_{n+1} = \frac{x^{n+1}}{(n+1)(n+2)} \).

Step 3: Calculate the Ratio

Calculate \( \frac{a_{n+1}}{a_n} = \frac{x^{n+1} (n(n+1))}{x^n ((n+1)(n+2))} = \frac{|x| n}{n+2} \). Simplifying, we get \( \left| \frac{a_{n+1}}{a_n} \right| = \frac{|x|}{n+2} \).

Step 4: Determine Convergence

The next step is to apply the limit as \( n \to \infty \): \( \lim_{{n \to \infty}} \frac{|x|}{n+2} = 0 \) for all \( x \). This implies the series converges for all \( x \), meaning it converges absolutely for all real numbers \( x \).

Step 5: Identify the Convergence Set

Since the limit of the ratio is 0 for all real \( x \), the convergence set of the power series is all real numbers. Therefore, the interval of convergence is \( (-\infty, \infty) \).

Key Concepts

Absolute Ratio Test

The Absolute Ratio Test is a common tool used in calculus to determine the convergence of a power series. It's particularly helpful as it gives us a way to handle series with variable terms. The test works by examining the limit of the absolute value of the ratio of consecutive terms in the series.For a series \( \sum a_n \), you calculate the ratio \( \left| \frac{a_{n+1}}{a_n} \right| \). If the limit of this ratio as \( n \to \infty \) is less than 1, then the series converges absolutely. This process involves simplifying the terms, computing the absolute ratio, and evaluating the limit:- Simplifying each term's expression.- Taking the limit of the ratio as \( n \) becomes infinitely large.- Comparing the result with 1.In our example, the terms are structured in such a way that the simplification leads to a ratio of \( \frac{|x|}{n+2} \) which approaches 0 as \( n \to \infty \). This signifies convergence for every possible value of \( x \), because 0 is certainly less than 1.

Convergence Set

The convergence set is all the values of \( x \) for which the power series converges. Discovering this set often follows the use of a convergence test, like the Absolute Ratio Test.Once the ratio test is applied and simplified as in the example above, where the \( \lim_{{n \to \infty}} \frac{|x|}{n+2} = 0 \), the implication is that the series converges.- Since the ratio yields 0, which is always less than 1 regardless of \( x \).- This insight allows us to conclude that the series is absolutely convergent for all real numbers. Thus, the series' convergence set is \( (-\infty, \infty) \), meaning no restrictions on \( x \), and the series converges for all real numbers.

General Term Identification

Identifying the general term in a power series is the critical first step in analyzing the series. This involves recognizing and expressing the recurring pattern or formula that describes every term in the series.In the given problem, the series is expressed as \( \frac{x}{1 \cdot 2}-\frac{x^{2}}{2 \cdot 3}+\frac{x^{3}}{3 \cdot 4}-\frac{x^{4}}{4 \cdot 5} \ldots \), indicating an alternating sign and a general structure.- Each term incorporates a power of \( x \), captured as \( x^n \).- There is a division by \( n(n+1) \), in the denominator, revealing a recognizable pattern. Therefore, the general term can be formalized as \( a_n = \frac{(-1)^{n+1} x^n}{n(n+1)} \). This step of pinpointing the general term allows for any subsequent convergence testing and evaluating the series' behavior.