Problem 17 - 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 \(\lim _{n \rightarrow \infty} a_{n}\) \(a_{n}=\frac{\ln n}{\sqrt{n}}\)

Step-by-Step Solution

Answer

The first five terms are 0, 0.490, 0.635, 0.693, 0.720; the sequence converges to 0.

Step 1: Substitute for n = 1

Plug in the value of n = 1 into the formula \(a_{n}=\frac{\ln n}{\sqrt{n}}\). \[a_{1} = \frac{\ln 1}{\sqrt{1}} = \frac{0}{1} = 0.\]

Step 2: Calculate a_2

Substitute n = 2 into the formula.\[a_{2} = \frac{\ln 2}{\sqrt{2}} \approx \frac{0.693}{1.414} \approx 0.490.\]

Step 3: Calculate a_3

Substitute n = 3 into the formula.\[a_{3} = \frac{\ln 3}{\sqrt{3}} \approx \frac{1.099}{1.732} \approx 0.635.\]

Step 4: Calculate a_4

Substitute n = 4 into the formula.\[a_{4} = \frac{\ln 4}{\sqrt{4}} = \frac{1.386}{2} = 0.693.\]

Step 5: Calculate a_5

Substitute n = 5 into the formula.\[a_{5} = \frac{\ln 5}{\sqrt{5}} \approx \frac{1.609}{2.236} \approx 0.720.\]

Step 6: Determine Convergence or Divergence

Analyze the expression \(a_{n}=\frac{\ln n}{\sqrt{n}}\). As \(n\) approaches infinity, \(\ln n\) grows slower than \(\sqrt{n}\), so \(a_{n}\) approaches 0. Therefore, the sequence converges.

Step 7: Find the Limit

Since the sequence converges, calculate the limit:\[ \lim_{n \to \infty} a_{n} = \lim_{n \to \infty} \frac{\ln n}{\sqrt{n}} = 0.\]

Key Concepts

Convergence and Divergence

In mathematics, understanding sequences and their behavior as they progress is crucial. A sequence either converges to a number as its terms go on indefinitely or diverges, meaning it does not approach any particular value. For a sequence to converge, its terms must get increasingly closer to a specific value, termed the limit.

Analyzing convergence involves checking if the terms shrink around a fixed point. If the sequence diverges, it could go to infinity, oscillate, or behave erratically without settling down. In the exercise provided, the sequence is expressed as:

• \(a_n = \frac{\ln n}{\sqrt{n}}\)

To determine convergence or divergence, it's crucial to consider the behavior of the expression as \(n\) grows very large. For this particular sequence, although both the numerator (\(\ln n\)) and denominator (\(\sqrt{n}\)) increase, the denominator grows faster, causing the sequence to converge to 0.

Limits of sequences

Limits provide a formal way to describe the behavior of a sequence as it extends towards infinity. For sequences that converge, the limit is the value the sequence settles on when its terms extend indefinitely. Basically, it's the point where the terms of the sequence get closer and closer as \(n\) increases.

To find the limit of the specific sequence \(a_n = \frac{\ln n}{\sqrt{n}}\), consider the relative growth of \(\ln n\) and \(\sqrt{n}\). Despite both tending to infinity, \(\ln n\) increases slower than \(\sqrt{n}\), leading to:

• \( \lim_{n \to \infty} a_n = 0 \)

This demonstrates how the sequence approaches a unique point, indicating that it converges. Understanding limits helps in foreseeing the ultimate behavior of sequences, especially those involving complex functions like logarithms and roots.

Exponential and logarithmic functions

Exponential and logarithmic functions play significant roles in mathematics, especially within calculus. They are essential tools for tackling sequences and series. Let's explore what they are and how they interact in our sequence.

Consider the natural logarithm \(\ln n\), which is the inverse of the exponential function \(e^x\). It steadily increases as \(n\) becomes larger, but at a diminishing rate. This is important in analyzing sequences like \(a_n\), where \(\ln n\) influences the numerator. Despite its unending increase, the rate of growth wanes.

\(\sqrt{n}\), a radical function, illustrates square roots and similarly escalates but at a faster pace than \(\ln n\). When they appear together, \(\ln n\) tends to grow slower than \(\sqrt{n}\), confirming that the sequence \(\frac{\ln n}{\sqrt{n}}\) converges to 0. Harnessing the insight from these functions assists in portraying the precise nature of sequences and structures in mathematics.