Problem 33 - Thomas Calculus

Question

Which of the series in Exercises 13 46 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 the comparison test.

Step 1: Analyze the Series Form

The series given is \( \sum_{n=3}^{\infty} \frac{1/n}{(\ln n) \sqrt{\ln^2 n - 1}} \). Notice that the terms are positive for \( n \geq 3 \), and involve a combination of \( n \), \( \ln n \), and \( \sqrt{\ln^2 n - 1} \). We need to determine whether these terms become sufficiently small as \( n \) approaches infinity to allow the series to converge.

Step 2: Simplify the Terms

Consider simplifying the denominator: \( (\ln n) \sqrt{\ln^2 n - 1} \approx (\ln n) \sqrt{(\ln n)^2} = (\ln n) \cdot (\ln n) = (\ln n)^2 \) for large \( n \). So, for large \( n \), each term approximately becomes \( \frac{1}{n (\ln n)^2} \).

Step 3: Conduct the Comparison Test

We will compare this approximation to a known convergent series. Specifically, consider the series \( \sum \frac{1}{n (\ln n)^2} \). Use the integral test to check its convergence where the integral \( \int \frac{1}{x (\ln x)^2} \, dx \) from 3 to infinity is convergent. Thus, the series \( \sum \frac{1}{n (\ln n)^2} \) converges.

Step 4: Conclusion Using the Comparison Test

Based on the comparison test, since \( \sum_{n=3}^{\infty} \frac{1/n}{(\ln n) \sqrt{\ln^2 n - 1}} \lesssim \sum_{n=3}^{\infty} \frac{1}{n (\ln n)^2} \) and the latter series converges, the given series also converges.

Key Concepts

Comparison Test

The comparison test helps determine the convergence or divergence of a series by comparing it to another series whose behavior is already known. If you have a series whose convergence or divergence is unknown, comparing it to a series with known behavior can be very illuminating. In general, you can use:

• If the terms of your unknown series are less than or equal to the terms of a known convergent series from a certain point onward, then your series also converges.
• If the terms of your unknown series are greater than or equal to the terms of a known divergent series from a certain point onward, your series also diverges.

In the given exercise, the series \( \sum \frac{1/n}{(\ln n) \sqrt{\ln^2 n - 1}} \) was compared to \( \sum \frac{1}{n (\ln n)^2} \). Since the latter is a known convergent p-series (where \( p > 1 \)), our series is shown to converge by the comparison test.

Integral Test

The integral test is a fundamental tool for determining the convergence of series, especially those involving continuous, positive, decreasing functions. If you have a series \( \sum a_n \), and there is a function \( f(x) \) such that \( f(n) = a_n \) and \( f(x) \) is continuous, positive, and decreasing, then:

• If the integral \( \int f(x) \, dx \) from 1 (or another lower bound) to infinity converges, the series \( \sum a_n \) also converges.
• If the integral diverges, then the series \( \sum a_n \) also diverges.

In this exercise, the integral test was applied to \( \int \frac{1}{x (\ln x)^2} \, dx \). By evaluating this integral from 3 to infinity, it is found to converge, supporting the conclusion that the series \( \sum \frac{1}{n (\ln n)^2} \) converges, hence assisting in the convergence conclusion for the original series.

Logarithmic Functions

Logarithmic functions are important when dealing with series whose terms involve logarithms, as seen in this exercise. With the series \( \sum \frac{1/n}{(\ln n) \sqrt{\ln^2 n - 1}} \), you encounter natural logs \( \ln n \). Understanding these can simplify computations and help in approximations. Properties of logarithms include:

• \( \ln(ab) = \ln a + \ln b \)
• \( \ln(a/b) = \ln a - \ln b \)
• \( \ln(a^b) = b \cdot \ln a \)
• \( \ln e = 1 \)

In the problem, an approximation was made: \( \sqrt{\ln^2 n - 1} \approx \ln n \) for large \( n \). This simplification makes it easier to harness the integral test and comparison test in determining convergence.

Approximation of Series

Approximating terms in a series is often necessary to apply tests or simplify analysis. Sometimes the form of a series is complex, involving expressions like logarithmic or exponential functions. Simplifying or approximating helps us compare to known series and apply convergence tests.In this task, the series \( \sum \frac{1/n}{(\ln n) \sqrt{\ln^2 n - 1}} \) was approximated as \( \sum \frac{1}{n (\ln n)^2} \) for large \( n \). Here, the approximation involves recognizing that unless it affects convergence, terms like \( \sqrt{\ln^2 n - 1} \) can be treated as \( \ln n \), simplifying the series into a form that can be handled with standard convergence tests. This technique is a powerful strategy for dealing with series that might seem complex at first glance.