Problem 29 - Thomas Calculus

Question

Which of the series in Exercises \(17-56\) converge, and which diverge? Use any method, and give reasons for your answers. $$ \sum_{n=2}^{\infty} \frac{1}{\sqrt{n} \ln n} $$

Step-by-Step Solution

Answer

The series converges.

Step 1: Identify the Type of Series

The given series is \( \sum_{n=2}^{\infty} \frac{1}{\sqrt{n} \ln n} \). It is a series with terms that are positive for all \( n \ge 2 \). We will analyze its convergence.

Step 2: Choose a Test for Convergence

The series resembles the form \( \sum \frac{1}{n^p} \) where \( p > 0 \), which suggests the use of the Integral Test. This is apt since the function involved is continuous and positive for \( n \ge 2 \).

Step 3: Apply the Integral Test

We consider the integral \( \int_{2}^{\infty} \frac{1}{\sqrt{x} \ln x} \, dx \). If the integral converges, then the series converges too, and vice versa.

Step 4: Simplify and Evaluate the Integral

Let \( I = \int_{2}^{\infty} \frac{1}{\sqrt{x} \ln x} \, dx \). Set \( u = \ln x \), thus \( du = \frac{1}{x} \, dx \) and \( dx = x \, du \). And note \( x = e^u \), so \( \sqrt{x} = e^{u/2} \).This transforms the function:\(I = \int_{\ln 2}^{\infty} \frac{1}{e^{u/2}} \, du\) Now, integrate:\(I = \int_{\ln 2}^{\infty} e^{-u/2} \, du\)The antiderivative of \( e^{-u/2} \) is \( -2e^{-u/2} \). Evaluate:\(= \left[-2e^{-u/2} \right]_{\ln 2}^{\infty} = 0 - \left(-2e^{-(\ln 2)/2}\right)\)\(= 2e^{-(\ln 2)/2} = 2 \times \frac{1}{\sqrt{2}}\)\(= \sqrt{2}\).Since the integral converges, the series converges.

Key Concepts

Integral Test

The Integral Test is a powerful tool that helps determine whether a series converges or diverges. It is particularly helpful when dealing with series that resemble a continuous and positive function, like the one from the exercise, which is positive for all \(n \geq 2\). To apply the Integral Test, you transform the series into an improper integral:

• You need a function \(f(x)\) that is continuous, positive, and decreasing for \(x \geq a\).
• The series \(\sum_{n=a}^{\infty} a_n\) and the integral \(\int_{a}^{\infty} f(x) \, dx\) are directly connected. If the integral converges, so does the series, and if it diverges, the series does too.

In the case of our exercise, we consider the integral \(\int_{2}^{\infty} \frac{1}{\sqrt{x} \ln x} \, dx\), which involves transforming the series terms into this improper integral form to test for convergence.

Positive Series

A positive series is a series where each term is positive, i.e., all terms \(a_n > 0\). This property simplifies the convergence analysis because certain tests, like the Integral Test, are directly applicable. With positive terms, we can avoid complex oscillations that might complicate convergence assessments.
For the exercise at hand, because \(\frac{1}{\sqrt{n} \ln n} > 0\) for all \(n \geq 2\), our task to determine convergence focuses solely on the size and growth of the series terms rather than any alternating signs. This makes it perfectly suited for techniques like the Integral Test that rely on positive functions.

Infinite Series

An infinite series is a sum of infinitely many terms, expressed as \(\sum_{n=a}^{\infty} a_n\). Convergence or divergence of an infinite series rests on whether the sum approaches a finite value or grows indefinitely.

• The series from the exercise, \(\sum_{n=2}^{\infty} \frac{1}{\sqrt{n} \ln n}\), is an infinite series under examination.
• The purpose of testing convergence is to see if adding infinitely many terms results in a finite sum.

Understanding infinite series and their convergence requires familiarity with several tests, such as the Integral, Comparison, and Ratio Tests, each suitable for different types of series configurations.

Antiderivative Evaluation

To evaluate an integral, finding the antiderivative (an inverse process to differentiation) is crucial. For the exercise, we compute the integral \(\int \frac{1}{e^{u/2}} \, du\), which requires a substitution method to simplify the expression.Here, the substitution \(u = \ln x\) transforms the integral into an easier form, because it results in an exponential function: \(e^{-u/2}\). Finding the antiderivative of this involves straightforward integration:

• The antiderivative of \(e^{-u/2}\) is \(-2e^{-u/2}\).
• Evaluate it from the limit at \(\ln 2\) to \(\infty\), resulting in a value of \(\sqrt{2}\).

Since the integral converges to a finite value, the original series converges as well, demonstrating the connection between antiderivative evaluation and series convergence.