Question

Use the Integral Test to determine the convergence of the given series. $$\sum_{n=2}^{\infty} \frac{1}{n \ln n}$$

Short answer

The series \( \sum_{n=2}^{\infty} \frac{1}{n \ln n} \) diverges.

Step-by-step solution

Identify the function for the Integral Test

The series given is \( \sum_{n=2}^{\infty} \frac{1}{n \ln n} \). To apply the Integral Test, we consider the function \( f(x) = \frac{1}{x \ln x} \) for \( x \geq 2 \), which is positive, continuous, and decreasing.

Set up the improper integral

The next step is to evaluate the improper integral \( \int_{2}^{\infty} \frac{1}{x \ln x} \, dx \). This integral will help us determine whether the corresponding series converges.

Make a substitution to simplify the integral

Use the substitution \( u = \ln x \), which implies \( du = \frac{1}{x} \, dx \). This transforms the integral limits: when \( x = 2\), \( u = \ln 2 \), and as \( x \to \infty \), \( u \to \infty \). Thus, \( \int_{2}^{\infty} \frac{1}{x \ln x} \, dx \) becomes \( \int_{\ln 2}^{\infty} \frac{1}{u} \, du \).

Evaluate the integral

Integrate \( \int \frac{1}{u} \, du \), which equals \( \ln |u| + C \). The improper integral becomes \( \left[ \ln |u| \right]_{\ln 2}^{\infty} = \ln(\infty) - \ln(\ln 2) = \infty \).

Conclude the result using the Integral Test

Since the improper integral \( \int_{2}^{\infty} \frac{1}{x \ln x} \, dx \) diverges, by the Integral Test, the series \( \sum_{n=2}^{\infty} \frac{1}{n \ln n} \) also diverges.

Key concepts

Convergence of Series

Understanding the convergence of series is crucial for analyzing whether a sum will approach a certain value as more terms are added. A series is simply a sum of terms in a sequence, written as \( \sum_{n=1}^{\infty} a_n \). To determine if this sum has a finite limit, we analyze its convergence. If the series converges, the sum approaches a finite number. If it diverges, the sum does not settle towards a finite value.

Convergence can be tested using various methods, and the Integral Test is one popular tool. It applies to positive, continuous, and decreasing functions. By comparing a series with a related improper integral, we can infer the series' behavior based on the integral's convergence or divergence.

• If the related integral converges, the series converges.
• If the integral diverges, the series diverges as well.

Understanding this connection provides a powerful method for tackling convergence questions.

Improper Integrals

Improper integrals are used to evaluate infinite intervals or functions with infinite discontinities. When evaluating them, such integrals take the form \( \int_{a}^{\infty} f(x) \, dx \) or may approach infinity at some point within the interval. These integrals extend the idea of a standard integral by allowing the limits of integration to include infinity.

In the context of the Integral Test, we often set up an improper integral to determine if a related series converges. For example, with the series \( \sum_{n=2}^{\infty} \frac{1}{n \ln n} \), we consider the improper integral \( \int_{2}^{\infty} \frac{1}{x \ln x} \, dx \). This involves an infinite upper limit, making the integration improper.

• An improper integral converges if it approaches a finite value.
• An improper integral diverges if it increases indefinitely.

Solving improper integrals often involves techniques like substitutions to simplify the evaluation and find out their behavior.

Substitution Method

The substitution method is an essential technique in calculus to simplify complex integrals by transforming them into more manageable forms. It is used by changing variables to simplify the function you're integrating. For instance, substitution can turn an inconvenient integral into one that is straightforward to evaluate.

In our exercise, we used the substitution \( u = \ln x \) to handle the integral \( \int_{2}^{\infty} \frac{1}{x \ln x} \, dx \). This transformation made the integral easier to evaluate since it simplified the function to \( \frac{1}{u} \). The direct integration of \( \int \frac{1}{u} \, du \) results in \( \ln |u| + C \), which is straightforward to compute.

• Choose a substitution that simplifies the integrand.
• Change the limits of integration according to the substitution.
• After solving, substitute back to the original variable if necessary.

Applying the substitution method effectively reduces the complexity of solving improper integrals and assists in determining the convergence of the series.