Problem 31 - Thomas Calculus

Question

Which of the series in Exercises \(11-40\) converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series' convergence or divergence.) $$ \sum_{n=3}^{\infty} \frac{(1 / n)}{(\ln n) \sqrt{\ln ^{2} n-1}} $$

Step-by-Step Solution

Answer

The series converges by comparison to the convergent series \( \sum \frac{1}{n (\ln n)^2} \).

Step 1: Identify the Type of Series

The series at hand is \( \sum_{n=3}^{\infty} \frac{(1 / n)}{(\ln n) \sqrt{\ln ^{2} n - 1}} \). This series involves logarithmic terms, suggesting a comparison test may be useful, possibly with an integral test, due to the complexity of the terms.

Step 2: Simplify the Term

Consider simplifying the term \( \frac{1/n}{(\ln n) \sqrt{(\ln n)^2 - 1}} \). This can be expressed as \( \frac{1}{n (\ln n) \sqrt{(\ln n)^2 - 1}} \). The denominator grows significantly as \( n \to \infty \), making the series akin to a challenging p-series or improper integral.

Step 3: Comparison Test - Choose a Series for Comparison

For large \( n \), \( \ln^2 n - 1 \approx \ln^2 n \), so \( \sqrt{\ln^2 n - 1} \approx \ln n \). Hence, \( \frac{1}{n (\ln n) \sqrt{\ln^2 n - 1}} \) is roughly \( \frac{1}{n (\ln n)^2} \). We can compare this to the series \( \sum \frac{1}{n (\ln n)^2} \).

Step 4: Analyze the Comparison Series - Integral Test

Consider \( \int_{3}^{\infty} \frac{1}{x (\ln x)^2} \, dx \). Performing the substitution \( u = \ln x \), \( du = \frac{1}{x} \, dx \), transforms the integral into \( \int \frac{1}{u^2} \, du \), which converges as it evaluates to \(-\frac{1}{u} = -\frac{1}{\ln x}\). Calculate \( \lim_{b \to \infty} \left[-\frac{1}{\ln x}\right]_{3}^{b} \). This limit evaluates to a finite value.

Step 5: Conclude Convergence of the Original Series

Since \( \sum \frac{1}{n (\ln n)^2} \) converges by the integral test, and our original series \( \sum \frac{1}{n (\ln n) \sqrt{\ln^2 n - 1}} \) is smaller than or similar to this convergent series for sufficiently large \( n \), our original series also converges by the direct comparison test.

Key Concepts

Comparison Test

The Comparison Test is a handy tool when trying to determine if a series converges or diverges. The idea is simple: you compare your series with another series whose convergence behavior you already know.

Here's how it works:

• Suppose you have two series, \(\sum a_n \)\ and \(\sum b_n \)\, where all the terms are positive.
• If you find that \(0 \leq a_n \leq b_n\)\ for all sufficiently large \(n\)\, and \(\sum b_n \)\ converges, then \(\sum a_n \)\ also converges.
• If \(a_n \geq b_n \geq 0\)\ and \(\sum b_n \)\ diverges, then \(\sum a_n \)\ also diverges.

In our exercise, we compared the series \(\sum \frac{1}{n (\ln n) \sqrt{\ln^2 n - 1}}\) with \(\sum \frac{1}{n (\ln n)^2}\), noting that the latter converges. By the comparison test, if our original series is smaller than the convergent series for large \(n\), it must also converge.

Integral Test

Sometimes series involve complex terms that make them tricky to analyze directly, especially when logarithmic and radical functions are involved. That's where the Integral Test can be valuable. The test connects the behavior of a series with an improper integral. The idea is simple: if you can integrate a function that you relate to the terms of the series, you can determine the convergence or divergence of the series.

Here's a step-by-step approach:

• Take the terms of your series, say \(a_n\), and let \(f(x)=a_x\) be a continuous, positive, and decreasing function for \(x \geq n\).
• Consider the improper integral \(\int_{n}^{\infty} f(x) \, dx\).
• If the integral converges, then the series \(\sum a_n\) converges as well.
• If the integral diverges, then the series diverges too.

In this exercise, we evaluated the integral \(\int_{3}^{\infty} \frac{1}{x (\ln x)^2} \, dx\). With a substitution method, we found it converges, confirming the convergence of the series \(\sum \frac{1}{n (\ln n)^2}\).

Logarithmic Functions

Logarithmic functions are a crucial concept in understanding the terms of our series. In mathematics, a logarithm answers the question: "To what exponent must we raise a specific base to obtain a particular number?"

Key properties of logarithmic functions that help simplify series:

• Logarithmic functions grow very slowly compared to polynomial functions.
• The natural logarithm, denoted as \(\ln n\), is commonly used in mathematical series.
• When \(\ln n\) increases, so does \(\ln^2 n\), making terms in our series shrink.

In our series, we encountered \(\ln n\) and \(\ln^2 n\) in the denominator, signifying that the terms shrink rapidly as \( n\) grows large. The slow growth of logarithmic functions in the denominator was a key reason the series could converge.

Infinite Series

An infinite series can be thought of as the sum of an infinite sequence of terms. It is crucial to determine whether such series converge (approach a specific value) or diverge (increase indefinitely without bound). For a series \(\sum a_n\), its convergence depends on whether the partial sums \(\sum_{i=1}^{n} a_i\) approach a finite limit as \(n\) becomes infinitely large.

Key considerations include:

• The terms \(a_n\) must tend toward zero as \(n\) increases for potential convergence.
• Different tests, like the Comparison Test and Integral Test, help decide if the series converges.

In the exercise, the infinite series \(\sum_{n=3}^{\infty} \frac{(1 / n)}{(\ln n) \sqrt{\ln ^{2} n-1}}\) involved rapid shrinking terms, supported by logarithmic properties, indicating convergence through the rigorous tests discussed.