Problem 3 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the convergence set for the given power series. $$ \sum_{n=1}^{\infty} \frac{x^{n}}{n^{2}} $$

Step-by-Step Solution

Answer

The convergence set is \(-1 \leq x \leq 1\).

Step 1: Identify the Type of Series

The given series is a power series, represented as \( \sum_{n=1}^{\infty} \frac{x^{n}}{n^{2}} \). Here, each term is of the form \( a_n x^n \) where \( a_n = \frac{1}{n^2} \). This is a real power series in \( x \).

Step 2: Apply the Ratio Test

To find the radius of convergence, we use the Ratio Test. Consider two consecutive terms of the series: \( a_n x^n = \frac{x^n}{n^2} \) and \( a_{n+1} x^{n+1} = \frac{x^{n+1}}{(n+1)^2} \). The Ratio Test requires us to evaluate \( \lim_{n \to \infty} \left| \frac{a_{n+1} x^{n+1}}{a_n x^n} \right| \).

Step 3: Simplify the Ratio Expression

Calculate the ratio: \[\left| \frac{a_{n+1} x^{n+1}}{a_n x^n} \right| = \left| \frac{x^{n+1}}{(n+1)^2} \cdot \frac{n^2}{x^n} \right| = \left| x \right| \cdot \left( \frac{n^2}{(n+1)^2} \right).\]This simplifies further to \( \left| x \right| \cdot \left( \frac{n^2}{(n+1)^2} \right) \).

Step 4: Evaluate the Limit

Evaluate the limit as \( n \to \infty \):\[\lim_{n \to \infty} \left| x \right| \cdot \left( \frac{n^2}{(n+1)^2} \right) = \left| x \right| \cdot \lim_{n \to \infty} \left( \frac{n^2}{(n+1)^2} \right) = \left| x \right| \cdot 1 = \left| x \right|.\]According to the Ratio Test, the series converges if \( \left| x \right| < 1 \).

Step 5: Determine the Convergence Set

From the evaluation of the Ratio Test, we conclude that the power series converges when \( \left| x \right| < 1 \), i.e., \( -1 < x < 1 \). For \( x = 1 \) or \( x = -1 \), the series should be checked separately. Using the Alternating Series Test or by considering the behavior of the series \( \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \) and \( \sum_{n=1}^{\infty} \frac{1}{n^2} \), both converge. Thus, the convergence set is \( -1 \leq x \leq 1 \).

Key Concepts

Ratio Test

The Ratio Test is a handy tool used to determine the convergence of a power series. It works by examining the ratio of consecutive terms in a series. In our exercise, we look at the series \( \sum_{n=1}^{\infty} \frac{x^{n}}{n^{2}} \) and focus on two consecutive terms: \( \frac{x^{n+1}}{(n+1)^2} \) and \( \frac{x^{n}}{n^2} \).

The Ratio Test requires us to find:

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

Using this formula, we simplify to get \( \left| x \right| \cdot \left( \frac{n^2}{(n+1)^2} \right) \). As \( n \to \infty \), the fraction \( \frac{n^2}{(n+1)^2} \) approaches 1, leaving us with \( \left| x \right| \).

For the series to converge using the Ratio Test, this last expression must be less than 1. Hence,

• \( \left| x \right| < 1 \)

This indicates that the series is convergent within this range.

Radius of Convergence

The radius of convergence is a key aspect of a power series' behavior. It defines an interval around a central value in which the series converges.

In our series \( \sum_{n=1}^{\infty} \frac{x^{n}}{n^{2}} \), the variable \( x \) determines if the series will converge. After applying the Ratio Test, we determined that

• \( \left| x \right| < 1 \)

This gives us a radius of convergence of 1. This means the series converges for values of \( x \) within an open interval from \(-1\) to \(1\). Values within this interval are part of the convergence set. The endpoints of the interval, \( x = 1 \) and \( x = -1 \), need further examination.

Alternating Series Test

This test is used to check the convergence of series where terms alternate in sign. For endpoint evaluation in our power series, let’s consider \( x = -1 \).

The series becomes \( \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \), which is an alternating series. It meets the criteria:

• The terms \( \frac{1}{n^2} \) decrease steadily.
• The limit of these terms as \( n \to \infty \) is zero.

Thus, by the Alternating Series Test, the series at \( x = -1 \) converges. Similarly, at \( x = 1 \), the series also converges as \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) is a known convergent series.

Limit Evaluation

Evaluating limits plays a pivotal role in ascertaining the convergence or divergence of a series. It is often used in conjunction with tests like the Ratio Test to define convergence properties.

As shown earlier, the limit:

• \( \lim_{n \to \infty} \left| x \right| \cdot \left( \frac{n^2}{(n+1)^2} \right) \)

simplifies to \( \left| x \right| \) after the ratio within the limit approaches 1.

This crucial step confirms the interval of convergence derived from the Ratio Test. Moreover, the limit ensures that the convergence is consistent across the interval \(-1 < x < 1\), establishing \( \left| x \right| < 1 \) as valid. Thus, our interval of convergence is bounded neatly by evaluating this limit.