Problem 8 - Thomas Calculus

Question

Use the Integral Test to determine if the series in Exercises \(1-10\) converge or diverge. Be sure to check that the conditions of the Integral Test are satisfied. $$ \sum_{n=2}^{\infty} \frac{\ln \left(n^{2}\right)}{n} $$

Step-by-Step Solution

Answer

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

Step 1: Verify Conditions for Integral Test

The Integral Test can be applied if the function \( f(x) = \frac{\ln(x^2)}{x} \) is positive, continuous, and decreasing for \( x \geq 2 \). \( f(x) \) is positive for \( x \geq 2 \) because both \( \ln(x^2) \) and \( x \) are positive for this range. \( \ln(x^2) \) is continuous as the composition of continuous functions, and \( x \) is continuous. To check if it’s decreasing, take the derivative of \( f(x) \).

Step 2: Compute the Derivative

Calculate \( f'(x) \) where \( f(x) = \frac{\ln(x^2)}{x} \). Using the quotient rule, \( f'(x) = \frac{d}{dx} \left( \frac{\ln(x^2)}{x} \right) = \frac{x \cdot \frac{d}{dx}(\ln(x^2)) - \ln(x^2) \cdot \frac{d}{dx}(x)}{x^2} = \frac{x \cdot \frac{2}{x} - \ln(x^2)}{x^2} = \frac{2 - \ln(x^2)}{x^2} \). For \( x \geq 2 \), \( 2 - \ln(x^2) \leq 0 \), indicating \( f(x) \) is decreasing.

Step 3: Set Up the Integral

Since the conditions are satisfied, apply the Integral Test. Evaluate the integral \( \int_{2}^{\infty} \frac{\ln(x^2)}{x} \, dx \), which simplifies to \( \int_{2}^{\infty} \frac{2\ln(x)}{x} \, dx \). This can be rewritten as \( 2 \int_{2}^{\infty} \frac{\ln(x)}{x} \, dx \).

Step 4: Evaluate the Integral

Use substitution: let \( u = \ln(x) \), then \( du = \frac{1}{x}dx \). The integral becomes \( 2 \int_{\ln(2)}^{\infty} u \, du \), which evaluates to \( 2 [\frac{u^2}{2}]_{\ln(2)}^{\infty} = [u^2]_{\ln(2)}^{\infty} \).

Step 5: Determine Divergence or Convergence

Evaluate \( [u^2]_{\ln(2)}^{\infty} = \infty - (\ln(2))^2 \), which diverges because the upper limit is infinite. Since the improper integral diverges, the series \( \sum_{n=2}^{\infty} \frac{\ln(n^2)}{n} \) also diverges by the Integral Test.

Key Concepts

Series Convergence

When dealing with series, we often need to determine whether they converge or diverge. Convergence means that as you add more and more terms, the total approaches a certain finite number. Divergence, on the other hand, implies that the total can keep increasing indefinitely without settling toward any particular value.
For a series like \( \sum_{n=2}^{\infty} \frac{\ln(n^2)}{n} \), using tests like the Integral Test helps us decide if the series converges or diverges. The Integral Test is useful because it allows us to compare a series to an improper integral. If the integral converges, so does the series; if the integral diverges, the series does too. This direct comparison simplifies the evaluation process considerably.
The series given in the exercise is analyzed by first ensuring that we can apply the Integral Test, which requires checking certain conditions for the related function is positive, continuous, and decreasing for the desired range.

Improper Integrals

Improper integrals extend the idea of definite integrals to functions with infinite limits or discontinuities in the range of integration. Evaluating an improper integral involves taking a limit to account for the infinity or discontinuity involved.
Taking the integral \( \int_{2}^{\infty} \frac{\ln(x^2)}{x} \, dx \) from the original exercise, we reformulate it using properties of logarithms and substitutions to find its exact behavior.
- If an integral leads to a finite number, it is considered convergent.
- If it results in an infinite value, it's divergent.
For the given integral, it diverges because as \( x \to \infty \), the logarithmic term increases without bound. Such divergence implies the series we are analyzing does not settle toward a finite value.

Calculus Derivatives

Understanding derivatives is crucial in many calculus concepts, including when determining if a function is decreasing. In the context of the Integral Test, we need to ensure the associated function does not increase for the test to be valid.
The derivation of \( f(x) = \frac{\ln(x^2)}{x} \) using the quotient rule is pivotal. The quotient rule tells us that if we have a function \( \frac{u(x)}{v(x)} \), the derivative is given by:\[ f'(x) = \frac{u'(x) v(x) - u(x) v'(x)}{v(x)^2} \]Applying it to our function, we're checking that the function is indeed decreasing for the range \( x \geq 2 \).
- A negative derivative implies that the function is decreasing.
The result \( f'(x) = \frac{2 - \ln(x^2)}{x^2} \) shows that this derivative is negative for \( x \geq 2 \), confirming the function is decreasing. This fact is essential for applying the Integral Test validly.