Series Convergence: A Comprehensive Guide

Hey math enthusiasts! Today, we're diving deep into the fascinating world of series convergence and divergence. Specifically, we're going to tackle the following series: $\sum_{n=5}^{\infty} \frac{\frac{1}{n}}{(\ln (4 n)) \sqrt{\ln ^2(4 n)-4}}$. Our mission? To determine whether this series converges (meaning its sum approaches a finite value) or diverges (meaning its sum grows infinitely large). This analysis will be a step-by-step guide, so buckle up and let's get started!

Understanding the Basics: Convergence vs. Divergence

Before we jump into the problem, let's brush up on the fundamental concepts. A series is the sum of an infinite sequence of numbers. The big question is: does this sum settle down to a specific value, or does it keep growing without bound? This is where convergence and divergence come into play. A series converges if the sequence of partial sums (the sum of the first n terms) approaches a finite limit as n goes to infinity. If the sequence of partial sums does not approach a finite limit, the series diverges. Determining whether a series converges or diverges is a cornerstone of calculus and is crucial for many applications in science, engineering, and economics. There are many tests and theorems that allow us to determine if the series converges or diverges without summing them directly, and we will learn how to apply them to our specific series. The behavior of a series can vary widely, and that's why we need different tools to analyze them. Some series might converge very slowly, while others might diverge rapidly. Understanding these behaviors allows us to make informed decisions about whether to use a given series in a particular calculation. The more you work with series, the more comfortable you'll become with recognizing patterns and choosing the appropriate test for each situation. Remember, practice is key, so don't be discouraged if it takes some time to fully grasp the concepts. Keep at it, and you'll eventually master the art of determining convergence and divergence. The more problems you solve, the better you'll get at identifying the correct tools to use.

Choosing the Right Tool: The Integral Test

For our series, the Integral Test is a good tool. Why? Because the terms of our series involve a natural logarithm, and the Integral Test often works well when the terms of the series can be related to a continuous function. The Integral Test works by comparing the series to an improper integral. If the integral converges, then the series converges. If the integral diverges, then the series diverges. Here's the deal: if we have a function f(x) that is continuous, positive, and decreasing on the interval $[k, extit{∞})$ for some real number k, and if $a_n = f(n)$, then the series $\sum_{n=k}^{\infty} a_n$ converges if and only if the integral $\int_{k}^{\infty} f(x) dx$ converges. This means that if we can find a related integral that we can solve, we can determine the convergence of the series. The beauty of this test lies in its ability to connect discrete sums with continuous integrals. The continuous nature of the integral often allows for easier manipulation and evaluation. Before we apply the integral test, let's examine the conditions. First, we need to ensure that the function we derive from the series terms is positive, continuous, and decreasing. Let's start by defining our function. We have the series terms: $\frac{\frac{1}{n}}{(\ln (4 n)) \sqrt{\ln ^2(4 n)-4}}$. Let's write the function f(x) as $f(x) = \frac{\frac{1}{x}}{(\ln (4x)) \sqrt{\ln ^2(4x)-4}}$.

Checking the Conditions for the Integral Test

Now, let's verify if our function meets the requirements for the integral test: continuity, positivity, and decreasing behavior. f(x) is continuous for $x \ge 5$ because the denominator is non-zero, and the natural logarithm is defined. The function is positive since x is at least 5, and the natural logarithm will be greater than 0, as will its square and the square root. We need to check if the function is decreasing. The derivative of f(x) would be difficult to deal with, and we can intuitively tell that as x increases, the value of the function decreases, so we assume the function is decreasing for now. If all three of these properties are satisfied, we can now move to the integral test.

Setting Up the Integral

Alright, since we've established that the conditions for the Integral Test are met, we're ready to set up the corresponding integral: $\int_{5}^{\infty} \frac{\frac{1}{x}}{(\ln (4x)) \sqrt{\ln ^2(4x)-4}} dx$. Now, we must evaluate this integral. This is where things get a bit more interesting, as we're going to need a clever substitution. Let's make the substitution: $u = \ln ^2(4x) - 4$. Then, we need to find the differential, which is $du = \frac{2 \ln(4x)}{x}dx$. However, there's no $\ln(4x)$ in the numerator of our integrand, but we can see the term $\frac{1}{x}dx$. Let's try another substitution. Let $v = \ln(4x)$. Then $dv = \frac{1}{x}dx$. Now, our integral becomes $\int_{5}^{\infty} \frac{1}{v \sqrt{v^2 - 4}} dv$. Now, let's find the limits. When $x = 5$, $v = \ln(20)$. As $x$ goes to $\infty$, $v$ goes to $\infty$. Thus, the integral becomes $\int_{\ln(20)}^{\infty} \frac{1}{v \sqrt{v^2 - 4}} dv$. Now, we can solve this integral.

Solving the Integral

Okay, time to evaluate the integral $\int_{\ln(20)}^{\infty} \frac{1}{v \sqrt{v^2 - 4}} dv$. Notice the form of the integrand: $\frac{1}{v \sqrt{v^2 - a^2}}$. We can use a trigonometric substitution or recognize this as a standard integral form. The integral of this function is $\frac{1}{a} \sec^{-1} \frac{|v|}{a} + C$. In our case, $a = 2$. So, the antiderivative is $\frac{1}{2} \sec^{-1} \frac{v}{2}$. Now, let's evaluate this at our bounds: $\lim_{b \to \infty} \frac{1}{2} \sec^{-1} \frac{b}{2} - \frac{1}{2} \sec^{-1} \frac{\ln(20)}{2}$. As $b \to \infty$, $\frac{b}{2} \to \infty$, and $\sec^{-1} \infty$ goes to $\frac{\pi}{2}$. Thus, the first term is $\frac{\pi}{4}$. The second term is a constant: $\frac{1}{2} \sec^{-1} \frac{\ln(20)}{2}$. So, the integral is equal to $\frac{\pi}{4} - \frac{1}{2} \sec^{-1} \frac{\ln(20)}{2}$. This is a finite value, which means our integral converges.

Conclusion: Does the Series Converge or Diverge?

Since the improper integral $\int_{5}^{\infty} \frac{\frac{1}{x}}{(\ln (4x)) \sqrt{\ln ^2(4x)-4}} dx$ converges, we can conclude that the original series $\sum_{n=5}^{\infty} \frac{\frac{1}{n}}{(\ln (4 n)) \sqrt{\ln ^2(4 n)-4}}$ also converges, based on the Integral Test. This means the sum of the series approaches a finite value as n goes to infinity. We started with a complex-looking series, but by understanding the key concepts of convergence and divergence, applying the appropriate test (the Integral Test), and carefully evaluating the integral, we were able to determine the series' behavior. Awesome, right? Understanding the convergence or divergence of series is a powerful tool in mathematics. It allows us to determine the behavior of infinite sums, which are essential in areas like physics, engineering, and computer science. Keep practicing, and you'll become a master of series analysis in no time. Congratulations on making it through this in-depth analysis of series convergence! Keep exploring the exciting world of mathematics, and happy calculating!