Problem 19 - Thomas Calculus

Question

Which of the series in Exercises \(11-40\) converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series' convergence or divergence.) $$ \sum_{n=2}^{\infty} \frac{\ln n}{n} $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Series

We are given the series \( \sum_{n=2}^{\infty} \frac{\ln n}{n} \). This is an infinite series and we need to determine its convergence or divergence.

Step 2: Apply the Integral Test

The Integral Test is applicable since \( f(n) = \frac{\ln n}{n} \) is positive, continuous, and decreasing for \( n \geq 2 \). We compare the series with the integral \( \int_{2}^{\infty} \frac{\ln x}{x} \, dx \).

Step 3: Set Up the Integral

We set up the integral \( \int_{2}^{\infty} \frac{\ln x}{x} \, dx \). To solve it, we use substitution. Let \( u = \ln x \), then \( du = \frac{1}{x} \, dx \), and the integral becomes \( \int \ln x \, d(\ln x) \).

Step 4: Integrate by Parts

To solve \( \int \ln x \, d(\ln x) \), we use integration by parts with \( u = \ln x \) and \( dv = \frac{1}{x} \, dx \), which gives \( du = \frac{1}{x} \, dx \) and \( v = x \). Thus, \( \int u \, dv = uv - \int v \, du \).

Step 5: Solve the Integral

Applying integration by parts, we have: \[ \int \ln x \, d(\ln x) = \ln x \cdot x - \int x \cdot \frac{1}{x} \, dx = x \ln x - \int 1 \, dx = x \ln x - x + C \]. The limits of integration for the improper integral are from 2 to \( \infty \). As \( x \to \infty \), \( x \ln x - x \to \infty \).

Step 6: Determine Convergence or Divergence

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

Key Concepts

Integral Test

The Integral Test is a valuable tool when determining the convergence or divergence of a series. It applies to series where the terms can be represented by a function that is positive, continuous, and decreasing. In our example, the series is \( \sum_{n=2}^{\infty} \frac{\ln n}{n} \), where the function can be written as \( f(x) = \frac{\ln x}{x} \). By checking the prerequisites:

• The function is positive for \( x \geq 2 \).
• It is continuous since both \( \ln x \) and \( x \) are continuous for \( x > 0 \).
• It is decreasing for \( x \geq 2 \) because the derivative \( \frac{d}{dx} \left( \frac{\ln x}{x} \right) \) is negative in this interval.

This confirms that the Integral Test is appropriate. By comparing the series to the integral \( \int_{2}^{\infty} \frac{\ln x}{x} \, dx \), we establish whether the series converges or diverges. If this integral converges, so does the series. Conversely, if the integral diverges, the series does as well.

Infinite Series

Infinite series are summations of an infinite sequence of terms. They can converge (sum to a finite value) or diverge (grow indefinitely without reaching a limit). For instance, the series \( \sum_{n=2}^{\infty} \frac{\ln n}{n} \) is an infinite series. Determining convergence involves comparing the series to known convergent or divergent series or applying tests like the Integral Test.
The series in question is particularly interesting as its terms decrease and involve both logarithmic and linear components. Such series are frequently challenging due to their irregular decay rates. However, understanding the behavior of the terms as \( n \to \infty \) is essential. For example:

• If terms of the series trend towards zero quickly, the series may converge.
• If they decrease too slowly, the series likely diverges.

In conclusion, seeing if the sum "fires off" to infinity or not, helps us determine if it is a convergent or divergent series.

Integration by Parts

Integration by parts is a fundamental technique used to find integrals. It is derived from the product rule for differentiation. Given problems with products of functions like \( \int u \, dv \), we use the formula:\[ \int u \, dv = uv - \int v \, du \]In the given series \( \sum_{n=2}^{\infty} \frac{\ln n}{n} \), to find \( \int \frac{\ln x}{x} \, dx \), integration by parts is necessary. Here's a breakdown of its application:

• Choose \( u = \ln x \) making \( du = \frac{1}{x} \, dx \).
• Set \( dv = \frac{1}{x} \, dx \), rendering \( v = x \).
• Substitute into the formula to get \( \ln x \cdot x - \int 1 \, dx \).
• Resulting in \( x \ln x - x + C \).

By evaluating this expression at the limits from 2 to \( \infty \), we observe divergence. Therefore, integration by parts not only simplifies integrals but directly aids in applying the Integral Test, thus helping to conclude that the series \( \sum_{n=2}^{\infty} \frac{\ln n}{n} \) diverges.