Problem 53 - Single Variable Calculus

Question

Use the Integral Test to show that \(\sum_{n=3}^{\infty} \frac{1}{n(\ln n)[\ln (\ln n)]^{p}}\) converges if \(p>1\) and diverges if \(p \leq 1\).

Step-by-Step Solution

Answer

Using the Integral Test, we find that the function \(f(n) = \frac{1}{n(\ln n)[\ln(\ln n)]^{p}}\) is continuous, positive, and decreasing for \(n \geq 3\). Evaluating the integral of the function over the range \([3, \infty)\), the integral converges if \(p > 1\) and diverges if \(p \leq 1\). Therefore, the series \(\sum_{n=3}^{\infty} \frac{1}{n(\ln n)[\ln(\ln n)]^{p}}\) converges if \(p > 1\) and diverges if \(p \leq 1\).

Step 1: Examine the conditions for the Integral Test

To perform the Integral Test, we must determine if the function \(f(n) = \frac{1}{n(\ln n)[\ln(\ln n)]^{p}}\) is continuous, positive, and decreasing for \(n \geq 3\). 1. Continuity: The function is a composition and product of continuous functions (i.e., \(n\), \(\ln n\) and \(\ln(\ln n)\)) for \(n \geq 3\), so it's continuous for that range. 2. Positivity: Since \(n \geq 3\), \(\ln n > 0\) and \(\ln(\ln n) > 0\), so the function is always positive. 3. Decreasing: To determine if the function is decreasing, we can take its derivative and verify if it's negative for the given range. The derivative of \(f(n)\) can be found using the chain rule and product rule. Computing the derivative: \(f'(n) = - \frac{\ln(\ln n) + p(\ln n)}{n^2(\ln n)^2[\ln(\ln n)]^{2p - 1}}\) Since \(f'(n)\) is negative for \(n \geq 3\), the function is decreasing for this range. Now, we've satisfied the preconditions for the Integral Test.

Step 2: Evaluate the integral

Now we need to evaluate the integral of the function \(f(x)\) over the range \([3, \infty)\) to determine if it converges: \(\int_{3}^{\infty} \frac{1}{x(\ln x)([\ln(\ln x)]^p)} dx\) To solve this integral, we can use the substitution method. Let \(u = \ln(\ln x)\) and \(x = e^{e^u}\). Computing the derivative of \(x = e^{e^u}\), we get: \(dx = e^u e^{e^u} du\) Now, we'll substitute these expressions into the integral: \(\int_{3}^{\infty} \frac{1}{x(\ln x)([\ln(\ln x)]^p)} dx = \int_{\ln(\ln 3)}^{\infty} \frac{e^u e^{e^u}}{e^{e^u} e^u(u^p)} du\)

Step 3: Simplify the integral and determine convergence/divergence

The integral simplifies to: \(\int_{\ln(\ln 3)}^{\infty} \frac{1}{u^p} du\) This integral converges if \(p > 1\) and diverges if \(p \leq 1\). Therefore, by the Integral Test, the original series also converges if \(p > 1\) and diverges if \(p \leq 1\).

Key Concepts

Series Convergence

In mathematics, understanding when an infinite series converges is key. A series is said to converge if the sum of its terms approaches a finite number as more and more terms are added. In simple terms, convergence means that as you keep adding terms of the series, you inch closer and closer to a specific value and eventually settle there.
For instance, to test the convergence of \(\sum_{n=3}^{\infty} \frac{1}{n(\ln n)[\ln(\ln n)]^{p}}\), we use the Integral Test, which involves integrating the analogous function over an interval. Convergence typically requires the integrated function to approach a limit or become sufficiently small.

• A series converges if its corresponding function's integral from a certain point to infinity converges.
• If the integral reaches a specific finite number, the series converges.

In our exercise, the series converges when \( p > 1 \), as shown through evaluating its integral. This means as long as \( p \) is greater than 1, adding more and more terms leads the sum to a specific, fixed value.

Series Divergence

Opposite to convergence, a series diverges if it doesn't settle on a finite number. This means, as you add up more terms of the series, the sum keeps growing and potentially tends to infinity.
Divergence can be identified through various tests, with the Integral Test being one of the common methods.

• The Integral Test reveals divergence if the integral of the function over the interval does not approach a finite limit.
• If the integral grows indefinitely, the corresponding series does the same, hence diverging.

In our case, the series \(\sum_{n=3}^{\infty} \frac{1}{n(\ln n)[\ln(\ln n)]^{p}}\) diverges if \( p \leq 1 \). This means, for these values of \( p \), the sum of the series continues to increase without bounds, never stabilizing on a particular number.

Substitution Method

The substitution method is a powerful calculus technique for simplifying integrals. It involves replacing a complicated part of the integrand with a new variable to make the integral easier to handle.
In our exercise, to evaluate \(\int_{3}^{\infty} \frac{1}{x(\ln x)([\ln(\ln x)]^p)} dx\),the substitution method is used by setting \(u = \ln(\ln x)\). This transforms the variable \(x\) into a function of \(u\), notably turning the integral into a more manageable form.

• This approach helps in breaking down complex integrals into simpler ones.
• After substitution, it's essential to change limits and differentiate the substituted function to replace \(dx\) with a new term in terms of \(du\).

Substitution streamlines the integration process, revealing whether the integral, and thus the series, converges or diverges.

Continuity of Functions

Continuity is an essential property of functions, especially when dealing with integrals and series tests like the Integral Test. A function is continuous when there is no break, jump, or abrupt change in its graph.
For the Integral Test, it is critical that the function is continuous over the interval of interest because this ensures that integrating it over that interval is valid.

• In our exercise, the function \(f(n) = \frac{1}{n(\ln n)[\ln(\ln n)]^{p}}\) is continuous for \(n \geq 3\) because each part, \(n, \ln n, \ln(\ln n)\), is continuous for \(n \geq 3\).
• Continuity helps guarantee the correct application of calculus techniques, such as differentiation and integration, without unforeseen issues.

This continuity ensures the smooth application of the Integral Test to determine the series' behavior, ensuring reliable results whether it converges or diverges.