Question

Use the integral test to determine whether the following sums converge. \(\sum_{n=2}^{\infty} \frac{1}{n \ln ^{2} n}\)

Short answer

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

Step-by-step solution

Check Conditions for the Integral Test

The integral test requires that the function associated with the given series, in this case, \( f(x) = \frac{1}{x \ln^2 x} \), be positive, continuous, and decreasing for \( x \geq 2 \). All of these conditions hold true for our function since \( x \ln^2 x \) is positive for \( x > 1 \) and is continuous and decreasing for \( x \geq 2 \).

Set Up the Improper Integral

To apply the integral test, consider the improper integral \( \int_{2}^{\infty} \frac{1}{x \ln^2 x} \, dx \). If this integral converges, then the series \( \sum_{n=2}^{\infty} \frac{1}{n \ln^2 n} \) also converges.

Substitution

Use the substitution \( u = \ln x \), which simplifies the integration process. Then, \( du = \frac{1}{x} \, dx \). The limits change as follows: when \( x = 2 \), \( u = \ln 2 \), and as \( x \to \infty \), \( u \to \infty \). The integral now becomes \( \int_{\ln 2}^{\infty} \frac{1}{u^2} \ du \).

Evaluate the Integral

The integral \( \int_{\ln 2}^{\infty} \frac{1}{u^2} \, du \) can be computed as \( \left[ -\frac{1}{u} \right]_{\ln 2}^{\infty} \). Calculating the improper integral gives \( 0 - \left(-\frac{1}{\ln 2}\right) = \frac{1}{\ln 2} \), which is a finite value indicating that the integral converges.

Conclude Using the Integral Test

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

Key concepts

Series Convergence

Understanding series convergence is central to many areas of mathematics and calculus. A series is essentially the sum of the terms of a sequence. To determine if a series converges, we look at the behavior of these terms as they progress towards infinity. A series converges if the sum approaches a finite limit as the number of terms grows, while it diverges if it does not.

In the given exercise, we use the integral test, which is one of the methods to determine convergence. The integral test is applicable when the function derived from the sequence is continuous, positive, and decreasing. If such is the case, then the series converges if and only if the corresponding integral converges. This makes it a powerful tool for analyzing series where simpler tests do not easily apply.

In our example, the series to be tested using the integral test is given by:

• Series: \( \sum_{n=2}^{\infty} \frac{1}{n \ln^2 n} \)
• Function: \( f(x) = \frac{1}{x \ln^2 x} \)

Improper Integral

Improper integrals are a type of integral where the integrand becomes unbounded, or where the interval of integration is infinite. In these cases, normal integration techniques might not apply, and special attention is needed to evaluate them.

For tackling improper integrals, we often need to consider limits to account for endpoints tending to infinity or places where the function becomes unbounded. Specifically, when checking the convergence of the series \( \sum_{n=2}^{\infty} \frac{1}{n \ln^2 n} \), we set up the integral:

• \( \int_{2}^{\infty} \frac{1}{x \ln^2 x} \, dx \)

This integral is known as improper because one of its limits is infinity, making direct evaluation impossible without using limits.

For convergence, if this integral reaches a finite value, it is said to converge. Conversely, if it grows without bound, it diverges. Evaluating this type of integral often involves techniques like substitution to simplify the problem.

Substitution Method

The substitution method is a key technique in calculus and is particularly useful for solving integrals that are difficult to solve directly. This method simplifies the problem by changing variables, making the integral more manageable.

In the given exercise, let's consider the substitution \( u = \ln x \). This choice is advantageous because it simplifies the integrand \( \frac{1}{x \ln^2 x} \) drastically:

• This substitution leads to \( du = \frac{1}{x} \, dx \), simplifying the original complex form.
• The limits of integration also change. Initially, when \( x = 2 \), \( u = \ln 2 \), and as \( x \to \infty \), \( u \to \infty \).

Applying this transformation, our integral becomes:

• \( \int_{\ln 2}^{\infty} \frac{1}{u^2} \, du \)

This new form is significantly easier to evaluate. Solving it from \( \ln 2 \) to infinity gives us a straightforward way to determine convergence based on whether or not the integral reaches a finite limit.