Problem 9 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 7–12, show that each series converges absolutely. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{n}{2^{n}} $$

Step-by-Step Solution

Answer

The series converges absolutely.

Step 1: Identify the Series

The given series is \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{n}{2^{n}} \). This is an alternating series because it includes \((-1)^{n+1}\), which alternates the sign of each term.

Step 2: Clarify Absolute Convergence

A series \( \sum a_n \) converges absolutely if the series \( \sum |a_n| \) converges. Here, \( a_n = (-1)^{n+1} \frac{n}{2^{n}} \), so \( |a_n| = \frac{n}{2^{n}} \). We need to determine if \( \sum_{n=1}^{\infty} \frac{n}{2^{n}} \) converges.

Step 3: Apply the Ratio Test

To test absolute convergence, apply the Ratio Test: if \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \), the series converges. Compute:\[\frac{a_{n+1}}{a_n} = \frac{n+1}{2^{n+1}} \cdot \frac{2^n}{n} = \frac{n+1}{2n}\]

Step 4: Evaluate the Limit

Calculate the limit of \( \frac{n+1}{2n} \) as \( n \to \infty \):\[\lim_{n \to \infty} \frac{n+1}{2n} = \lim_{n \to \infty} \left( \frac{n}{2n} + \frac{1}{2n} \right) = \frac{1}{2} + 0 = \frac{1}{2}\]Since \( \frac{1}{2} < 1 \), the series \( \sum \frac{n}{2^{n}} \) converges.

Step 5: Conclusion of Absolute Convergence

Since the series of absolute values \( \sum \frac{n}{2^{n}} \) converges by the Ratio Test, the original series \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{n}{2^{n}} \) converges absolutely.

Key Concepts

Alternating Series

An alternating series is a series whose terms alternate in sign. This means, for every term, the sign changes between positive and negative. Let's look at an example of an alternating series:

• The series \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{n}{2^n} \) is alternating because it has the term \((-1)^{n+1}\), which switches the sign for every \(n\).
• Alternating series can still converge even if individual terms keep switching from positive to negative.

One common method to determine convergence of alternating series is the Alternating Series Test. However, in the case of absolute convergence, we apply different tests like the Ratio Test, as we'll discuss in a later section.

Ratio Test

The Ratio Test is a powerful tool used to determine the convergence or divergence of a series. It evaluates the ratio of successive terms in the series and examines its limit as \(n\) approaches infinity. Here’s how it works:

• For a series \(\sum a_n\), consider the ratio \(\frac{a_{n+1}}{a_n}\).
• If \(\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| < 1\), the series converges absolutely.
• If the limit is greater than 1, the series diverges, and if it equals 1, the test is inconclusive.

In our specific example, the given series is \(\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{2^n}\). To apply the Ratio Test, we first examine the absolute values, thus considering \(\sum_{n=1}^{\infty} \frac{n}{2^n}\). By determining the limit \(\lim_{n \to \infty} \frac{n+1}{2n} = \frac{1}{2}\), we find it is less than 1, confirming absolute convergence.

Convergent Series

A convergent series is one where the sum of its terms approaches a finite number as more terms are added. Several methods can determine series convergence, each with specific requirements:

• For a series to be absolutely convergent, the series of absolute values must converge. That is, \(\sum |a_n|\) must approach a finite limit.
• The given series \(\sum_{n=1}^{\infty}(-1)^{n+1} \frac{n}{2^n}\) is proven to converge absolutely by showing \(\sum_{n=1}^{\infty} \frac{n}{2^n}\) itself converges using the Ratio Test.

Absolute convergence is particularly strong because it implies regular convergence, providing a stable foundation for evaluating series behavior. By proving absolute convergence, we gain confidence in the series’ behavior and the reliability of the sum’s finite limit.