Problem 17 - 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. $$ 1-\frac{x}{1 \cdot 3}+\frac{x^{2}}{2 \cdot 4}-\frac{x^{3}}{3 \cdot 5}+\frac{x^{4}}{4 \cdot 6}-\cdots $$

Step-by-Step Solution

Answer

The convergence set of the series is \((-1, 1)\).

Step 1: Identify the General Term

First, observe the given power series and identify the pattern. The series is:\[1 - \frac{x}{1 \cdot 3} + \frac{x^2}{2 \cdot 4} - \frac{x^3}{3 \cdot 5} + \frac{x^4}{4 \cdot 6} - \cdots\]The general term \(a_n\) can be observed as:\[a_n = \frac{(-1)^n x^n}{n (n+2)}\].

Step 2: Apply the Absolute Ratio Test

The Absolute Ratio Test is used to determine the convergence of the power series. This involves calculating the limit:\[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]Plug in the general term formula:\[ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(-1)^{n+1} x^{n+1} / \left((n+1)(n+3)\right)}{-1^n x^n / \left(n(n+2)\right)} \right| = \left| \frac{x(n(n+2))}{(n+1)(n+3)} \right|\].

Step 3: Simplify the Ratio Expression

Simplify the expression:\[\left| \frac{x(n(n+2))}{(n+1)(n+3)} \right| = \left| x \right| \cdot \frac{n^2 + 2n}{n^2 + 4n + 3}\]Now take the limit as \(n\) approaches infinity:\[L = \lim_{n \to \infty} \left| x \right| \cdot \frac{n^2 + 2n}{n^2 + 4n + 3} = \left| x \right| \cdot 1 = \left| x \right|\].

Step 4: Determine the Radius of Convergence

For the series to converge, by the Ratio Test, the limit \(L\) must be less than 1:\[\left| x \right| < 1\].This inequality defines the interval of convergence. Thus, the radius of convergence \( R = 1 \).

Step 5: Check Endpoints for Convergence

The interval of convergence is initially \((-1, 1)\) from the ratio test, and we must test the endpoints.1. If \( x = 1 \): The series is \[1 - \frac{1}{1 \cdot 3} + \frac{1}{2 \cdot 4} - \frac{1}{3 \cdot 5} + \cdots \], a divergent alternating series.2. If \( x = -1 \): The series is \[1 + \frac{1}{1 \cdot 3} + \frac{1}{2 \cdot 4} + \frac{1}{3 \cdot 5} + \cdots \], a divergent series, as terms remain positive and do not decrease.Hence, the interval of convergence remains \((-1, 1)\).

Key Concepts

General Term

When dealing with power series, it's crucial to identify the general term. Observing a pattern within the series helps us construct a general formula for any term. In the provided series:

• Both the sign and the numerator follow a predictable pattern.
• The denominator features products of consecutive natural numbers incremented by two.

For our series, the general term \( a_n \) is expressed as: \[ a_n = \frac{(-1)^n x^n}{n(n+2)} \] This formula:

• Includes \((-1)^n\), which alternates the signs of the terms, giving us the alternating series.
• Uses \(x^n\) for the powers of \(x\) as seen in traditional power series.
• Has a denominator \(n(n+2)\), which evolves as this specific series progresses.

Absolute Ratio Test

To determine if our power series converges, we apply the Absolute Ratio Test. This method checks whether the limit of the ratio of successive terms is less than 1. Here's how it works for our series:

• We start by forming the ratio \(\frac{a_{n+1}}{a_n}\), substituting the general expressions discovered earlier.

After simplifying, the key component left to evaluate is:\[\lim_{n \to \infty} \left| \frac{x(n(n+2))}{(n+1)(n+3)} \right| = \left| x \right|\]In this step:

• The complex parts of the function essentially cancel out, allowing \(|x|\) to become the focal point.

If this \(|x|\) is less than 1, the Ratio Test tells us that the series converges.

Radius of Convergence

With the Absolute Ratio Test, we reach a crucial conclusion related to convergence called the "radius of convergence." It provides a clear boundary within which the series converges:

• From the simplified ratio, \(|x| < 1\) ensures convergence.
• This inequality directly suggests that the radius of convergence, \(R\), is 1.

This radius implies:

• The series will converge for any value of \(x\) within a distance of 1 from the origin.
• Mathematically, this means \(-1 < x < 1\).

Interval of Convergence

The interval of convergence tells us the precise range of \(x\) values for which our series converges. Initially determined from the radius, this is refined by examining the endpoints:

• If \(x = 1\), the series becomes divergent due to alternating but not diminishing terms.
• For \(x = -1\), similar non-decaying positive terms lead to divergence.

Hence, testing endpoints confirms the interval remains: \((-1, 1)\).

• This interval indicates exact values for which the entire series remains valid and convergent.

It's important because it sets real-world application boundaries of a given mathematical model or theory.