Question

Use the Integral Test to determine the convergence of the given series. $$\sum_{n=1}^{\infty} \frac{\ln n}{n^{3}}$$

Short answer

The series \( \sum_{n=1}^{\infty} \frac{\ln n}{n^{3}} \) converges by the Integral Test.

Step-by-step solution

Identify the Function

To apply the Integral Test, first identify the continuous, positive, and decreasing function. In this case, consider the function \( f(x) = \frac{\ln x}{x^3} \), where \( x \geq 1 \).

Verify Conditions for the Integral Test

Check if \( f(x) = \frac{\ln x}{x^3} \) is positive, continuous, and decreasing for \( x \geq 1 \). The function is positive since logarithm and polynomial terms are positive for \( x \geq 1 \). As \( x \) increases, the denominator grows faster than the numerator, so the function is decreasing.

Set Up the Integral

The Integral Test requires evaluating the improper integral \( \int_{1}^{\infty} \frac{\ln x}{x^3} \, dx \). We will determine if this integral converges.

Evaluate the Integral

To evaluate \( \int_{1}^{\infty} \frac{\ln x}{x^3} \, dx \), use integration by parts. Let \( u = \ln x \) and \( dv = \frac{1}{x^3} \, dx \). Then \( du = \frac{1}{x} \, dx \) and \( v = -\frac{1}{2x^2} \).

Apply Integration by Parts

Using integration by parts, \[ \int u \, dv = uv - \int v \, du \]. Substitute \( u \) and \( v \) to get: \[-\frac{\ln x}{2x^2} \Bigg|_{1}^{\infty} + \frac{1}{2} \int_{1}^{\infty} \frac{1}{x^3} \, dx \].

Evaluate the Boundary Terms

Evaluate each term:- \(-\frac{\ln(\infty)}{2(\infty)^2} = 0\) because \(\frac{\ln x}{x^2} \to 0\) as \( x \to \infty \).- \(-\frac{\ln(1)}{2(1)^2} = 0\).

Simplify Remaining Integral

Now find \( \frac{1}{2} \int_{1}^{\infty} \frac{1}{x^3} \, dx \). This simplifies to \( \frac{1}{2} \cdot \left[ -\frac{1}{2x^2} \right]\Bigg|_{1}^{\infty} \).

Evaluate Remaining Integral

Compute \( \frac{1}{2} \left( 0 - (-\frac{1}{2}) \right) = \frac{1}{4} \). Since this integral converges, the series converges by the Integral Test.

Key concepts

Convergence of Series

In mathematics, determining the convergence or divergence of a series is crucial in understanding its behavior as the number of terms goes to infinity. A series is simply the sum of the terms in a sequence. Imagine you keep adding smaller and smaller numbers together forever. Does it approach a particular number (converge), or does it keep getting larger or oscillate wildly (diverge)? This is what we mean by convergence.When a series converges, the terms are getting closer to a specific limit. The series in the exercise is represented as:\[\sum_{n=1}^{\infty} \frac{\ln n}{n^{3}}\]It involves a logarithmic term and a polynomial term. It's essential to analyze such a series properly using specific tests like the Integral Test to confirm its convergence.Understanding how to evaluate the convergence helps in many areas in mathematics and in practical situations like calculating probabilities and solving differential equations.

Integration by Parts

Integration by parts is a useful technique for solving integrals, especially when dealing with products of functions. This method is grounded on the product rule for differentiation and can be stated as:\[\int u \, dv = uv - \int v \, du\]In our exercise, to evaluate \( \int \frac{\ln x}{x^3} \, dx \), we first need to choose our \( u \) and \( dv \). Let's set \( u = \ln x \) and \( dv = \frac{1}{x^3} \, dx \). By differentiating and integrating, respectively, we find:

• \( du = \frac{1}{x} \, dx \)
• \( v = -\frac{1}{2x^2} \)

Now, we can apply the formula. Each step should be taken with care to ensure all calculations align, especially when approaching the limits of an improper integral. Integration by parts allows us to transform and simplify complex integrals for easier evaluation.

Improper Integrals

An improper integral is one where at least one of the limits of integration is infinite or where the function to be integrated becomes infinite within the limits of integration. These types of integrals often appear when using the Integral Test for convergence of series.In the given exercise, our integral is:\[\int_{1}^{\infty} \frac{\ln x}{x^3} \, dx\]Notice that the upper limit is \( \infty \), which makes it an improper integral. Such integrals require a limit process to evaluate, typically by taking the limit as \( b \to \infty \) for:\[\int_{1}^{b} \frac{\ln x}{x^3} \, dx\]This approach ensures that we accurately determine whether the integral converges or diverges. In calculus, improper integrals help extend the usefulness of the integral concept beyond standard finite boundaries.

Decreasing Function

A decreasing function is one that consistently becomes smaller as the input value increases. Mathematically, a function \( f(x) \) is decreasing on an interval if for any two numbers \( x_1 \) and \( x_2 \) in that interval, if \( x_1 < x_2 \), then \( f(x_1) \geq f(x_2) \).For the Integral Test, having a decreasing function is an essential criterion. The function from our exercise is:\[f(x) = \frac{\ln x}{x^3}\]For \( x \geq 1 \), the denominator \( x^3 \) grows faster than the numerator \( \ln x \), confirming that \( f(x) \) decreases as \( x \) increases. Verification of this behavior is key in applying the Integral Test accurately.By confirming that \( f(x) \) is decreasing, we can confidently use the integral test to check for convergence, simplifying our work as we evaluate such series.