Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(\sum_{n=1}^{\infty} \frac{\ln n}{2^{n}}\)

Step-by-Step Solution

Answer

The series converges by the Ratio Test.

Step 1: Understanding the Series

The given series is \( \sum_{n=1}^{\infty} \frac{\ln n}{2^{n}} \). This is an infinite series, where each term is of the form \( \frac{\ln n}{2^n} \). We want to determine whether this series converges or diverges.

Step 2: Applying the Ratio Test

To determine the convergence of the series, we apply the Ratio Test. The Ratio Test requires us to evaluate the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \), where \( a_n = \frac{\ln n}{2^n} \).

Step 3: Setting Up the Terms for the Ratio Test

Calculate \( a_{n+1} = \frac{\ln(n+1)}{2^{n+1}} \). Therefore, the ratio \( \frac{a_{n+1}}{a_n} \) becomes \( \frac{\ln(n+1)}{2^{n+1}} \times \frac{2^n}{\ln n} = \frac{\ln(n+1)}{2 \ln n} \).

Step 4: Computing the Limit

Now compute the limit \( L = \lim_{n \to \infty} \frac{\ln(n+1)}{2 \ln n} \). Note that as \( n \to \infty \), \( \ln(n+1) \approx \ln n \). Therefore, the limit can be simplified to \( \frac{1}{2} \lim_{n \to \infty} \frac{\ln(n+1)}{\ln n} = \frac{1}{2} \).

Step 5: Concluding the Test

The Ratio Test states that if \( L < 1 \), the series converges. Here, \( L = \frac{1}{2} < 1 \), so the series \( \sum_{n=1}^{\infty} \frac{\ln n}{2^n} \) converges.

Key Concepts

Ratio Test

The Ratio Test is a popular method in calculus to determine the convergence of series. It involves calculating the limit of the absolute value of the ratio of consecutive terms.
The process goes as follows:

• Consider a series with terms denoted as \(a_n\). The Ratio Test examines the limit \(L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
• If \(L < 1\), then the series converges absolutely.
• If \(L > 1\) or if the limit is infinite, the series diverges.
• If \(L = 1\), the test is inconclusive.

In the exercise, applying the Ratio Test to the series \(\sum_{n=1}^{\infty} \frac{\ln n}{2^n}\) resulted in \(L = \frac{1}{2}\), confirming the convergence.

Infinite Series

An infinite series is essentially the sum of infinitely many terms, and mathematical techniques help determine if it approaches a certain value (converges) or grows without bound (diverges).
An infinite series is usually denoted as \(\sum_{n=1}^{\infty} a_n\), where each term follows a particular pattern.

• If the series converges, the sum reaches a finite limit as more terms are added.
• If the series diverges, the partial sums grow indefinitely without approaching a specific value.

Understanding whether an infinite series converges or diverges is crucial in calculus, impacting everything from integrals to advanced function analysis.

Natural Logarithm

The natural logarithm, often denoted as \(\ln(x)\), is a mathematical function that tells us the power to which the number \(e\) (approximately 2.71828) must be raised to obtain the value \(x\).
It is an essential concept in calculus, particularly in dealing with exponential growth processes.

• For small changes in \(x\), the natural logarithm can approximate how these changes influence functions involving exponentiation.
• Natural logarithms often appear in solving differential equations and analyzing growth models.

In the given exercise, \(\ln(n)\) is part of each term in the infinite series, which is crucial for applying the Ratio Test and checking for convergence.

Calculus

Calculus is the branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It provides powerful tools for analyzing changes.
Some foundational concepts include:

• Differential Calculus: Concerned with analyzing and understanding derivatives, which represent rates of change.
• Integral Calculus: Focuses on integrals, with applications in computing areas under curves and accumulated quantities.
• Series: Infinite series, such as the one in our exercise, are also a central part of calculus and are used to represent functions as limits of sums.

Through methods like the Ratio Test, calculus helps determine whether a series converges, giving insight into the behavior of functions.