Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(\frac{\ln 2}{2^{2}}+\frac{\ln 3}{3^{2}}+\frac{\ln 4}{4^{2}}+\frac{\ln 5}{5^{2}}+\cdots\)

Step-by-Step Solution

Answer

The series converges.

Step 1: Recognizing the Series

The given expression is an infinite series where each term is of the form \( \frac{\ln n}{n^2} \). This is a convergent series because the numerator \( \ln n \) grows much slower than the denominator \( n^2 \).

Step 2: Applying the Integral Test

To determine convergence, let's use the integral test. Consider the function \( f(x) = \frac{\ln x}{x^2} \), which is positive, continuous, and decreasing for \( x \geq 2 \).

Step 3: Setting Up the Integral

Evaluate the integral \( \int_{2}^{\infty} \frac{\ln x}{x^2} \, dx \). If the integral converges, so does the series.

Step 4: Integration by Parts

Use integration by parts to evaluate the integral. Let \( u = \ln x \) (therefore \( du = \frac{1}{x} \, dx \)) and \( dv = \frac{1}{x^2} \, dx \) (therefore \( v = -\frac{1}{x} \)).

Step 5: Applying Integration by Parts Formula

The integration by parts formula \( \int u \, dv = uv - \int v \, du \) leads to: \[ \int \frac{\ln x}{x^2} \, dx = -\frac{\ln x}{x} + \int \frac{1}{x^2} \, dx. \]

Step 6: Solving Remaining Integral

The remaining integral is \( \int \frac{1}{x^2} \, dx = -\frac{1}{x} \). So, \[-\frac{\ln x}{x} - \frac{1}{x}.\]

Step 7: Evaluating the Definite Integral

Evaluate \([-\frac{\ln x}{x} - \frac{1}{x}]\) from 2 to \(\infty\). As \(x \to \infty\), the expression approaches 0. Evaluating from 2 gives a finite value.

Step 8: Conclusion on Convergence

Since the integral \( \int_{2}^{\infty} \frac{\ln x}{x^2} \, dx \) is finite, the series \( \sum_{n=2}^{\infty} \frac{\ln n}{n^2} \) converges.

Key Concepts

Convergence

When faced with the task of determining if an infinite series converges, one of our primary concerns is whether the sum of all its terms adds up to a finite value. In our example, the series is comprised of terms of the form \( \frac{\ln n}{n^2} \). Here, the numerator \( \ln n \) indicates a slow growth rate, especially when compared to the quadratic growth of \( n^2 \) in the denominator.

For a series to converge, the terms must get progressively smaller as \( n \) increases to infinity. In this case, because as \( n \) increases, \( n^2 \) grows faster than \( \ln n \), the terms individually diminish towards zero. Thus, our infinite series does indeed converge.

It's worth noting that a mere decrease in term size doesn't guarantee convergence, but if the series passes certain tests like the integral test, it confirms the series converges. By understanding how quick growth rates contrast between numerators like \( \ln n \) and a denominator of \( n^2 \), convergence can be more easily predicted.

Integral Test

The integral test is a powerful tool we use to analyze the convergence of infinite series. The core of this test involves comparing the series to an improper integral. If the integral converges, so does the series; conversely, if the integral diverges, the series does too.

In our case, we apply the integral test by considering the function \( f(x) = \frac{\ln x}{x^2} \). This function is positive, continuous, and decreasing for \( x \geq 2 \), making it suitable for the test. The task is to evaluate the integral \( \int_{2}^{\infty} \frac{\ln x}{x^2} \, dx \).

Evaluating this improper integral will determine convergence. Since we've already concluded its convergence in the exercise, it assures us of the series' convergence. This approach gives us a more structured method compared to simply comparing term sizes and complements tests like the comparison test and ratio test.

Integration by Parts

Integration by parts is a technique from calculus especially useful when integrating products of functions, such as \( \frac{\ln x}{x^2} \). In our scenario, we choose \( u = \ln x \) and \( dv = \frac{1}{x^2} \, dx \). From these, we derive \( du = \frac{1}{x} \, dx \) and \( v = -\frac{1}{x} \).

This gives us the integration by parts formula: \( \int u \, dv = uv - \int v \, du \). Applying it to our integral, we have:\[ \int \frac{\ln x}{x^2} \, dx = -\frac{\ln x}{x} + \int \frac{1}{x^2} \, dx \]

The remaining integral, \( \int \frac{1}{x^2} \, dx \), can be evaluated as \( -\frac{1}{x} \). Thus, the solution becomes:\[-\frac{\ln x}{x} - \frac{1}{x} \]

With this expression, evaluating the integral from 2 to \( \infty \) provides a conclusive finite result, supporting the convergence of the original series. This technique simplifies the complexity involved in integrating fractional logarithmic functions, revealing clear results when combined skillfully.