Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-20, an explicit formula for \(a_{n}\) is given. Write the first five terms of \(\left\\{a_{n}\right\\}\), determine whether the sequence converges or diverges, and, if it converges, find \(\lim _{n \rightarrow \infty} a_{n}\) \(a_{n}=\frac{\ln (1 / n)}{\sqrt{2 n}}\)

Step-by-Step Solution

Answer

The sequence converges to 0, with the first five terms: 0, \( \frac{-\ln(2)}{2} \), \( \frac{-\ln(3)}{\sqrt{6}} \), \( \frac{-\ln(4)}{2\sqrt{2}} \), \( \frac{-\ln(5)}{\sqrt{10}} \).

Step 1: Identify the Expression for Terms

We are given the sequence formula \( a_n = \frac{\ln(1/n)}{\sqrt{2n}} \). Our task is to write the first five terms of the sequence, determine if the sequence converges or diverges, and find the limit if it converges.

Step 2: Calculate the First Five Terms

Calculate each term by substituting \(n = 1, 2, 3, 4, 5\) into the formula: - For \(n = 1\), \(a_1 = \frac{\ln(1/1)}{\sqrt{2 \cdot 1}} = 0 \)- For \(n = 2\), \(a_2 = \frac{\ln(1/2)}{\sqrt{2 \cdot 2}} = \frac{-\ln(2)}{2} \)- For \(n = 3\), \(a_3 = \frac{\ln(1/3)}{\sqrt{2 \cdot 3}} = \frac{-\ln(3)}{\sqrt{6}} \)- For \(n = 4\), \(a_4 = \frac{\ln(1/4)}{\sqrt{2 \cdot 4}} = \frac{-\ln(4)}{2\sqrt{2}} \)- For \(n = 5\), \(a_5 = \frac{\ln(1/5)}{\sqrt{2 \cdot 5}} = \frac{-\ln(5)}{\sqrt{10}} \)Thus, the first five terms are: 0, \(\frac{-\ln(2)}{2}\), \(\frac{-\ln(3)}{\sqrt{6}}\), \(\frac{-\ln(4)}{2\sqrt{2}}\), \(\frac{-\ln(5)}{\sqrt{10}}\).

Step 3: Analyze Convergence or Divergence

To determine the behavior of the sequence as \( n \to \infty \), we examine the limit of \( a_n \): \[ \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{\ln(1/n)}{\sqrt{2n}} = \lim_{n \to \infty} \frac{-\ln(n)}{\sqrt{2n}} \]As \( n \to \infty \), \( \ln(n) \) grows much slower than \( \sqrt{2n} \), causing the fraction to approach zero. Hence, the sequence converges to 0.

Step 4: Conclusion and Limit

Since \( \lim_{n \to \infty} \frac{-\ln(n)}{\sqrt{2n}} = 0 \), the sequence converges, and the limit is 0.

Key Concepts

Sequence Analysis

When assessing a sequence, we start by understanding its explicit formula and identifying the nature of its terms. In this case, our sequence is given by \( a_n = \frac{\ln(1/n)}{\sqrt{2n}} \). Our goal is to compute the initial terms, recognize patterns, and predict behavior as \( n \) becomes very large. Sequence analysis involves calculating individual terms by substituting integers into the formula given for \( n \). For example:

• For \( n = 1 \): The term is \( 0 \) because \( \ln(1) = 0 \).
• For \( n = 2 \): The term becomes \( \frac{-\ln(2)}{2} \).
• As \( n \) increases, we see the negative logarithm and radical terms affecting the sequence's behavior.

It's essential to observe if terms are shrinking or growing and whether they head toward a specific value (convergence), or keep growing or oscillating (divergence). Through sequence analysis, we gain critical insights about the nature and behavior of the sequence.

Limit Calculation

When discussing limits in sequences, our main goal is to determine what value the sequence approaches as \( n \) gets very large. For our sequence, the limit analysis involves:

• Recognizing that each term \( a_n = \frac{-\ln(n)}{\sqrt{2n}} \) represents a fraction.
• In this fraction, the numerator \( -\ln(n) \) grows slowly compared to the denominator \( \sqrt{2n} \), creating a diminishing fraction.

As \( n \) approaches infinity, we evaluate the limit:\[\lim_{n \to \infty} \frac{-\ln(n)}{\sqrt{2n}} = 0\]The key point here is that even though the logarithm grows with \( n \), it cannot keep up with the rate at which the square root increases. Hence, the sequence converges to the limit of zero.

Logarithmic Functions

Logarithms are critical in understanding some sequences due to their unique growth properties. In our sequence, the term \( \ln(1/n) \) can be rewritten as \( -\ln(n) \), highlighting the interplay between the logarithm and the decreasing nature of terms in the sequence:

• Logarithms grow slower than polynomial and exponential functions, which is why they appear in simplifications where their effect diminishes over large \( n \).
• A key property here is \( \ln(ab) = \ln(a) + \ln(b) \), which helps break down complex expressions.

Understanding logarithmic functions is crucial to predicting how the sequence's numerator evolves and influences the overall behavior. This aids in identifying the convergence to zero as seen in this sequence.

Roots and Radicals

Roots and radicals appear in our sequence as the denominator \( \sqrt{2n} \). They play a major role in determining how quickly the sequence terms decrease:

• Radicals, like square roots, tend to increase slower than their argument, \( n \), as \( n \) increases.
• The presence of \( \sqrt{2n} \) in the denominator affects the fraction by growing rapidly enough to overpower the slow-growing logarithmic numerator.

Thus, radicals help navigate towards zero for large \( n \) in sequences where their terms include logarithms in the numerators. Understanding radicals is key to solving limit problems where they appear, as they fundamentally alter the growth rate and ultimate behavior of expressions.