Problem 34

Question

Does $\sum_{n=3}^{\infty} 1 /[n \cdot \ln n \cdot \ln (\ln n)]$ converge or diverge? Explain.

Step-by-Step Solution

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Answer

The series diverges by the integral test.

1Step 1: Understand the Series

We are given the series $\sum_{n=3}^{\infty} \frac{1}{n \cdot \ln n \cdot \ln(\ln n)}$. This is an infinite series with terms dependent on function evaluations involving $n, \ln n,$ and $\ln(\ln n)$. Our task is to determine if it converges.

2Step 2: Apply the Integral Test

The series' expression suggests that the integral test could be used for convergence. For the integral test, we need to check the integral $\int_{3}^{\infty} \frac{1}{x \ln x \ln(\ln x)} \, dx$. If this improper integral converges, so does the series. If it diverges, so does the series.

3Step 3: Simplify the Integral

Consider the substitution $u = \ln(\ln x)$, which gives $du = \frac{1}{\ln x} \cdot \frac{1}{x} \, dx$. This substitution simplifies the integral to $\int \frac{1}{u} \, du$, which is easier to evaluate.

4Step 4: Evaluate the Integral

Upon integration, $\int \frac{1}{u} \, du = \ln|u| + C$. Substituting back $u = \ln(\ln x)$, we obtain $\ln|\ln(\ln x)|$. This suggests that the integral diverges as $x \to \infty$, since $\ln(\ln x)$ increases without bound, making $\ln|\ln(\ln x)|$ grow.

5Step 5: Conclude Using the Integral Test

Since the integral $\int_{3}^{\infty} \frac{1}{x \ln x \ln(\ln x)} \, dx$ diverges, by the integral test, the series $\sum_{n=3}^{\infty} \frac{1}{n \cdot \ln n \cdot \ln(\ln n)}$ also diverges.

Key Concepts

Integral TestImproper IntegralsSubstitution MethodDivergence

Integral Test

The integral test is a valuable tool in determining the convergence or divergence of an infinite series. It links a series with an associated improper integral. The basic idea is simple: if we have a series $ \sum_{n=a}^{\infty} a_n $, we can consider the function $ f(x) $ that matches $ a_n $ when evaluated at positive integer points

We need $ f(x) $ to be continuous, positive, and decreasing for $ x \geq a $.
If the integral $ \int_{a}^{\infty} f(x) \, dx $ converges, then the series $ \sum_{n=a}^{\infty} a_n $ converges.
If the integral diverges, so does the series.

This method is particularly useful because it sometimes simplifies the problem to evaluating an integral, which can be a more straightforward computational task.

Improper Integrals

Improper integrals arise in calculus when we try to compute the area under a curve that extends indefinitely. These integrals frequently appear in the context of determining series convergence using the integral test. Improper integrals can occur in two scenarios:

Infinite limits of integration, where the upper limit is infinity.
Unbounded functions, where the function approaches infinity at some point in the domain of integration.

In the example of our exercise, we dealt with the improper integral $ \int_{3}^{\infty} \frac{1}{x \ln x \ln(\ln x)} \, dx $, which had an infinite limit of integration. Solving such an integral involves evaluating limits to determine whether the integral converges to a finite number or diverges to infinity. Here, this improper integral diverged, indicating the same for the related series.

Substitution Method

The substitution method simplifies complex integrals by changing variables. This involves a strategic change to transform an integral into a more manageable form. Here's what you do:

Identify a part of the integrand that, when changed into a new variable, will simplify the integration process.
Choose a substitution function $ u = g(x) $. This choice should simplify the original function.
Derive $ du $ from the substitution, replacing the corresponding derivative $ dx $ part of the integral.
Rewrite the integral in terms of $ u $ and $ du $.
- Example substitution: Set $ u = \ln(\ln x) $.
- Then, $ du = \frac{1}{\ln x} \cdot \frac{1}{x} \, dx $.
Evaluate the new, simplified integral in terms of $ u $, then substitute back to $ x $ if necessary.

This method proved valuable in the exercise, simplifying the integral to $ \int \frac{1}{u} \, du $, which is much easier to solve.

Divergence

Divergence in the context of an infinite series means that the sum of the series' terms does not settle around a finite number. This can happen when the terms of a series continue to add up to infinity, rather than converging towards a particular value.In our specific exercise, divergence was proved using the integral test. Once the associated improper integral was found to diverge as $ x \to \infty $, we concluded that $ \sum_{n=3}^{\infty} \frac{1}{n \cdot \ln n \cdot \ln(\ln n)} $ also diverges. Divergence is crucial to understand:

It helps identify whether summing the terms will yield a meaningful, finite result.
It warns us when an infinite series does not behave nicely and rather grows infinitely large or oscillates.

Understanding when and why a series diverges helps in the analysis of series and their applications in different fields of mathematics.

Problem 34

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