Problem 7 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the Power Series

The given power series is \( \sum_{n=1}^{\infty}(-1)^{n} \frac{(x-2)^{n}}{n} \). This is a typical power series centered at \( x = 2 \) with the general term \( a_n = (-1)^{n} \frac{(x-2)^{n}}{n} \).

Step 2: Apply the Alternating Series Test

For convergence of an alternating series \( \sum_{n=1}^{\infty} (-1)^n a_n \), the series will converge if the sequence \( a_n \) is monotonically decreasing and \( \lim_{n \to \infty} a_n = 0 \). Here, \( a_n = \frac{|x-2|^n}{n} \).

Step 3: Determine the Behavior of \( |x-2|^n/n \)

As \( n \to \infty \), the term \( |x-2|^n/n \) tends to \( 0 \) for \( |x-2| < 1 \) due to the nature of exponential decay exceeding linear growth. Thus, for \( |x-2| < 1 \), the series converges.

Step 4: Analyze the Endpoint \( x = 3 \)

At \( x = 3 \), \( |x-2| = 1 \), so the series becomes \( \sum_{n=1}^{\infty} (-1)^{n} \frac{1}{n} \), which is the alternating harmonic series. This series is known to converge by the Alternating Series Test.

Step 5: Analyze the Endpoint \( x = 1 \)

At \( x = 1 \), \( |x-2| = 1 \), so the series becomes \( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n} \), which is another form of the alternating harmonic series. This series also converges by the Alternating Series Test.

Step 6: Conclude Convergence Set

Since the series converges for \( |x-2| < 1 \) and at both endpoints \( x = 3 \) and \( x = 1 \), the interval of convergence is \([1, 3]\).

Key Concepts

Alternating Series Test

The Alternating Series Test is an important tool used to determine the convergence of alternating series, which are series of terms that change sign between positive and negative. For a series of the form \( \sum_{n=1}^{\infty} (-1)^n a_n \), it can be shown to converge if two conditions are met:

• The sequence \( a_n \) is monotonically decreasing. This means each term is smaller than the preceding term.
• The limit of \( a_n \) as \( n \to \infty \) is zero, \( \lim_{n \to \infty} a_n = 0 \).

In the given power series \( \sum_{n=1}^{\infty} (-1)^n \frac{(x-2)^n}{n} \), the terms \( a_n \) are \( \frac{|x-2|^n}{n} \). Confirming these terms decrease with increasing \( n \) and disappear to zero ensures convergence by this test when \( |x-2| < 1 \).
Hence, alternating series like the harmonic series \( \sum \frac{(-1)^{n}}{n} \) often serve as classic examples where the Alternating Series Test is applicable.

Convergence Set

The convergence set of a power series is the set of all values for which the series converges. For the series \( \sum (-1)^n \frac{(x-2)^n}{n} \), determining where the series converges involves examining the behavior of \( |x-2| \).In the context of power series:

• A convergence set often takes the form of an interval within which the series behaves nicely, meaning it converges.
• Such series have a center, in this case, \( x = 2 \).
• The distance from the center, \( |x-2| \), determines if \( x \) is within this interval.

For this series, if \( |x-2| < 1 \), the series converges by the Alternating Series Test. Including the endpoints may also be required to determine complete convergence, as is revealed by additional analysis at \( x = 1 \) and \( x = 3 \).
The inclusion of these endpoints depends on the behavior of the series there.

Radius of Convergence

The radius of convergence determines how far from the center of the series the interval extends. It is found using the condition \( |x-2| < 1 \). Here:

• The radius is 1, derived from \( |x-2| < 1 \), as this forms an open interval \((1, 3)\) around the center \( x=2 \).
• Within this radius, the given series converges absolutely or conditionally.

To find the radius, consider the expression derived from the basic form of power series convergence, often utilizing the ratio test or the root test. However, for simplicity in this series, the Alternating Series Test directly helps render the condition \( |x-2| < 1 \).
This approach discriminates between points where convergence occurs and points that demand further investigation.

Endpoint Analysis

When determining the convergence set, endpoint analysis involves checking the behavior of the series at the boundaries of the convergence interval. For the series \( \sum_{n=1}^{\infty}(-1)^n \frac{(x-2)^n}{n} \):

• At its endpoints, check specifically at \( x=3 \) and \( x=1 \) to determine if convergence still holds.
• At \( x=3 \), the series simplifies to \( \sum (-1)^n \frac{1}{n} \), known to converge by the Alternating Series Test (it's the alternating harmonic series).
• Similarly, at \( x=1 \), the series \( \sum (-1)^{n+1} \frac{1}{n} \) also converges by the same logic.

Inclusion of both endpoints \( x=1 \) and \( x=3 \) in the convergence interval yields the closed interval \([1, 3]\). These calculations confirm whether the interval defined by the radius of convergence includes or excludes the endpoints, resulting in a comprehensive understanding of the convergence set.