Problem 26 - 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. $$ \frac{x-2}{1^{2}}+\frac{(x-2)^{2}}{2^{2}}+\frac{(x-2)^{3}}{3^{2}}+\frac{(x-2)^{4}}{4^{2}}+\cdots $$

Step-by-Step Solution

Answer

The convergence set is \((1, 3)\).

Step 1: Identify the General Term

Observe the given series and identify the general term. The series can be written as \( \sum_{n=1}^{\infty} \frac{(x-2)^n}{n^2} \). Here, the nth term \( a_n \) of the series is \( \frac{(x-2)^n}{n^2} \).

Step 2: Apply the Absolute Ratio Test

The Absolute Ratio Test involves finding the limit \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). Substitute \( a_n = \frac{(x-2)^n}{n^2} \) and \( a_{n+1} = \frac{(x-2)^{n+1}}{(n+1)^2} \) into the formula:\[ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(x-2)^{n+1}}{(n+1)^2} \cdot \frac{n^2}{(x-2)^n} \right| = \left| \frac{(x-2) \cdot n^2}{(n+1)^2} \right| \].

Step 3: Simplify the Limit Expression

Now, simplify \( \lim_{n \to \infty} \left| \frac{(x-2) n^2}{(n+1)^2} \right| \) further:\[ \left| x-2 \right| \lim_{n \to \infty} \frac{n^2}{(n+1)^2} = \left| x-2 \right| \lim_{n \to \infty} \left( \frac{n}{n+1} \right)^2 \]. Since \( \lim_{n \to \infty} \left( \frac{n}{n+1} \right)^2 = 1 \), the limit becomes \( \left| x-2 \right| \cdot 1 = \left| x-2 \right| \).

Step 4: Establish the Convergence Criterion

According to the Absolute Ratio Test, for the series to converge, the limit found in the previous step must be less than 1. So, we have:\( \left| x-2 \right| < 1 \).

Step 5: Determine the Convergence Set

Interpret the inequality \( \left| x-2 \right| < 1 \). This implies that the values of \( x \) are between 1 and 3. So, the interval of convergence is \( (1, 3) \). This is the set of \( x \) values for which the series converges.

Key Concepts

Absolute Ratio Test

The Absolute Ratio Test is a practical and reliable method used to determine the convergence of infinite series. When analyzing series, it’s often efficient to assess whether the series will converge or diverge by examining its terms. The Absolute Ratio Test relies on calculating the limit of the absolute ratio of consecutive terms in the series.
To apply the Absolute Ratio Test, consider a series with terms described as \( a_n \). For each consecutive pair of terms, \( a_n \) and \( a_{n+1} \), you compute the limit:

• \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \), also known as the ratio.

If this limit is less than 1, the series is guaranteed to converge. However, if the result is greater than 1, the series will diverge. Should the limit equal precisely 1, the test is inconclusive, and we might need to use other tests. This method is particularly helpful for series where the terms include factors that grow or shrink exponentially, such as factorials or exponential functions.

General Term Identification

Identifying the general term in a series is a crucial first step in understanding its behavior. A general term, often denoted as \( a_n \), represents a single element in the series, defined for all integers \( n \) beyond a starting point. Discovering the general term allows us to express any part of the series without listing all terms sequentially.
For our example power series, the general term \( a_n \) is given by \( \frac{(x-2)^n}{n^2} \). This compact formula provides a simple representation of each term in the infinite series. Recognizing this relationship is pivotal, as it lays the foundation for applying convergence tests, such as the Absolute Ratio Test. Through this understanding, you can explore the series' behavior in terms of both convergence and divergence.

Interval of Convergence

The interval of convergence is a range of values for the variable within a power series where the series converges. Understanding this interval is essential because it tells us exactly for which values of \( x \) the series behaves correctly—that is, where the sum of the infinite terms results in a finite number.
To find the interval, we start by applying the Absolute Ratio Test and solving any inequalities that arise. In our case, the inequality \( \left| x - 2 \right| < 1 \) establishes that the series converges. Solving this simple inequality, we obtain the interval \( (1, 3) \). This solution reveals that for any \( x \) lying between 1 and 3, the series will converge.

Convergence Criterion

The convergence criterion provides a mathematical basis for when a series converges based on conditions derived from specific tests. In our instance, using the Absolute Ratio Test set the stage for establishing when the series would converge. We specifically had to analyze the condition \( \left| x - 2 \right| < 1 \).
This criterion shows that for the series to be convergent, the distance between \( x \) and 2 must be less than 1, effectively setting \( 1 < x < 3 \) as the range of convergence. This technical requirement ensures that within this interval, the entire series aggregates to a finite sum, thus confirming its convergence. Recognizing this convergence criterion is essential as it guides how solutions are interpreted and brought into real-world applications, where series often model complex phenomena.