Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

Determine convergence or divergence for each of the series. Indicate the test you use. $$ \sum_{n=1}^{\infty} \frac{\ln n}{2^{n}} $$

Step-by-Step Solution

Answer

The series converges by the ratio test.

Step 1: Choose the Test

For the series \( \sum_{n=1}^{\infty} \frac{\ln n}{2^n} \), the ratio test is a suitable choice due to the presence of the exponential term \( 2^n \).

Step 2: Apply the Ratio Test

The ratio test involves calculating \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \), where \( a_n = \frac{\ln n}{2^n} \). First, identify \( a_{n+1} = \frac{\ln(n+1)}{2^{n+1}} \). Then compute:\[ \frac{a_{n+1}}{a_n} = \frac{\ln(n+1)}{2^{n+1}} \cdot \frac{2^n}{\ln n} = \frac{\ln(n+1)}{2 \ln n} \]

Step 3: Evaluate the Limit

Find \( \lim_{n \to \infty} \left( \frac{\ln(n+1)}{2 \ln n} \right) \). As \( n \to \infty \), both \( \ln(n+1) \) and \( \ln n \) tend toward infinity, and the dominant behavior can be analyzed using L'Hôpital's rule or recognizing that \( \ln(n+1) \sim \ln n \). Thus:\[ \lim_{n \to \infty} \frac{\ln(n+1)}{\ln n} = 1 \]Therefore, the limit is \( \lim_{n \to \infty} \frac{1}{2} = \frac{1}{2} \).

Step 4: Apply the Conclusion of the Ratio Test

Since \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \frac{1}{2} < 1 \), the ratio test indicates that the series \( \sum_{n=1}^{\infty} \frac{\ln n}{2^n} \) is absolutely convergent.

Key Concepts

Ratio Test

The Ratio Test is a powerful tool used to determine the convergence or divergence of an infinite series. It can be especially useful when dealing with series that include factorials, exponential terms, or powers. The essence of the Ratio Test involves examining the ratio of successive terms in a series:

• If the limit of the absolute value of the ratio of consecutive terms is less than 1, the series is absolutely convergent.
• If the limit is greater than 1, or infinite, the series diverges.
• If the limit is exactly 1, the test is inconclusive, and other methods must be used to determine the series' convergence.

Consider the series \( \sum_{n=1}^{\infty} \frac{\ln n}{2^n} \). To apply the Ratio Test, calculate:\[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\]where \( a_n = \frac{\ln n}{2^n} \). This helps verify if the series converges or not by showing that the limit is less than 1, which confirms convergence.

Infinite Series

An infinite series is essentially a sum of infinitely many terms. It is denoted as \( \sum_{n=1}^{\infty} a_n \), where \( a_n \) represents the general term of the series. These series can be thought of as a continual adding process:

• If the sum of all terms approaches a finite number as more terms are added, the series is said to converge.
• If adding more terms continues to increase the sum without bound, the series diverges.

Infinite series are foundational in calculus and are used to express functions, evaluate limits, and solve differential equations. The series \( \sum_{n=1}^{\infty} \frac{\ln n}{2^n} \) is an example of an infinite series, where understanding its behavior as \( n \to \infty \) is crucial for determining convergence or divergence.

Convergent Series

A convergent series is an infinite series whose terms approach a specific finite sum. This happens when, after adding an infinite number of terms, the series approaches a real limit. The Ratio Test is one of several tests used to check for convergence in series.To illustrate, consider the series \( \sum_{n=1}^{\infty} \frac{\ln n}{2^n} \). When using the Ratio Test, we find that:\[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \frac{1}{2} < 1\]Given that the limit is less than 1, the series is confirmed to be absolutely convergent. This means not only does the series converge, but it also converges absolutely, which is a stronger form of convergence. Understanding that a series converges means recognizing that its sum is finite and well-defined in the context of infinite terms.