Problem 5 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the General Term

The general term of the given power series is \((-1)^{n+1} \frac{x^n}{n^2}\). This identification is crucial for determining the radius of convergence and the interval of convergence.

Step 2: Apply the Ratio Test

To find the radius of convergence, apply the ratio test. Compute the limit \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_{n}} \right| = \lim_{n \to \infty} \left| \frac{(-1)^{n+2} x^{n+1}/(n+1)^2}{(-1)^{n+1} x^n/n^2} \right| \). Simplify the expression to get \( \lim_{n \to \infty} \left| x \right| \frac{n^2}{(n+1)^2} \).

Step 3: Simplify the Limit Expression

Further simplify the expression from Step 2: \( \lim_{n \to \infty} \left| x \right| \frac{n^2}{(n+1)^2} = \lim_{n \to \infty} \left( |x| \cdot \frac{1}{1 + \frac{2}{n} + \frac{1}{n^2}} \right) = |x| \). This is because \( \lim_{n \to \infty} \frac{n^2}{(n+1)^2} = 1 \).

Step 4: Determine the Radius of Convergence

According to the ratio test, for convergence, the limit \(|x|\) must be less than 1. Therefore, the radius of convergence \(R\) is 1.

Step 5: Find the Interval of Convergence

With the radius of convergence \(R = 1\), the power series converges when \(|x| < 1\). Examine the endpoints \(x = 1\) and \(x = -1\) separately to determine if they should be included in the interval.

Step 6: Test Convergence at x = 1

Substitute \(x = 1\) into the series: \( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n^2} \). This is an alternating series that converges because \( \frac{1}{n^2} \) is a decreasing sequence that approaches 0.

Step 7: Test Convergence at x = -1

Substitute \(x = -1\) into the series: \( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(-1)^n}{n^2} = \sum_{n=1}^{\infty} \frac{-1}{n^2} \). This series becomes \( \sum_{n=1}^{\infty} \frac{-1}{n^2} \) which diverges because the terms do not alternate in sign.

Step 8: Conclude the Interval of Convergence

The power series converges for \(|x| < 1\) and at \(x = 1\), but diverges at \(x = -1\). The convergence interval is therefore \([-1, 1)\).

Key Concepts

Ratio Test

The Ratio Test is a popular method to determine the convergence of power series. When you have a series \( \sum a_n \), you can apply the ratio test by finding the limit \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). This helps to assess whether the series converges or diverges by comparing successive terms.
In the context of power series, we focus on expressions of the form \( a_n = c_n x^n \). The ratio test helps establish a range of \( x \) values where the series behaves in a certain way.

• If the limit is less than 1, the series converges absolutely.
• If the limit is greater than 1, the series diverges.
• If the limit is exactly 1, the test is inconclusive.

These principles are used to derive the radius of convergence for a power series.

Radius of Convergence

Once you apply the ratio test to a power series, the next step is determining the radius of convergence, often denoted by \( R \). The radius of convergence is the distance within which the power series converges.
From the ratio test, we typically derive an expression such as \( \lim_{n \to \infty} \left| x \right| \frac{n^2}{(n+1)^2} = |x| \). This result is crucial as it starts determining valid \( x \) values. The series converges whenever \( |x| \lt R \), where \( R \) is determined as the value that makes the limit converge.
Here, the expression simplifies to \( |x| < 1 \) which tells us \( R = 1 \). The radius of convergence essentially outlines a circle in the complex plane within which convergence is guaranteed.

Interval of Convergence

Finding the interval of convergence for a power series involves not just knowing the radius of convergence but also testing the endpoints of the interval. When \( R \) is known, the general interval for convergence is then \( -R < x < R \).
Still, it's essential to check whether the series converges at the endpoints, \( x = R \) and \( x = -R \). In our example, at \( x = 1 \), the series converges due to the behavior as an alternating series. However, at \( x = -1 \), the series diverges.
Thus, when considering whether \( x = 1 \) or \( x = -1 \) is included, we determine that the interval of convergence for the series \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{x^{n}}{n^{2}} \) is \([-1, 1)\). This means that all \( x \) values within this interval, except \( x = -1 \), give us a convergent series.

Alternating Series Test

The Alternating Series Test provides a criterion for the convergence of series whose terms alternate in sign, such as \( \sum (-1)^n b_n \). For the test to confirm that the series converges, two conditions must be met:

• The absolute value of the terms \( b_n \) must steadily decrease.
• The limit of \( b_n \) as \( n \to \infty \) must be 0.

These checks confirm the convergence of a series that alternates and whose terms sequentially get smaller.
In our series, when testing for \( x = 1 \), we observe that the terms \( \frac{1}{n^2} \) meet both criteria: they decrease and approach zero. This is why the series converges. Conversely, at \( x = -1 \), the sequence \( (-1)^{n+1} \frac{(-1)^n}{n^2} \) does not alternate in a required fashion, leading it to diverge as established earlier.