Chapter 11: Problem 21
Use comparison tests to determine whether the infinite series converge or diverge. \(\sum_{n=1}^{x} \frac{\ln n}{n^{2}}\)
Short Answer
Expert verified
The series converges by the Comparison Test.
Step by step solution
01
Identify Series Type
The given series is \(\sum_{n=1}^{\infty} \frac{\ln n}{n^{2}}\). It is a series involving a term with a logarithmic function in the numerator and a power of \(n\) in the denominator.
02
Choose a Comparison Series
To use the comparison test, choose a simpler series that resembles the given one. We know that the series \(\sum_{n=1}^{\infty} \frac{1}{n^{2}}\) converges (this is a p-series with \(p=2 > 1\)). We will compare the given series with this convergent series.
03
Compare Terms
For large \(n\), the logarithmic function grows slowly compared to polynomial functions, meaning \(\frac{\ln n}{n^{2}}\) is similar to \(\frac{1}{n^{2}}\) but smaller. Specifically, \(\ln n < n\) for \(n \geq 3\), hence \(\frac{\ln n}{n^{2}} < \frac{n}{n^{2}} = \frac{1}{n}\).
04
Apply Limit Comparison Test
Compute the limit of the ratio of the terms from the given series and the chosen p-series: \[\lim_{n \to \infty} \frac{\frac{\ln n}{n^{2}}}{\frac{1}{n^{2}}} = \lim_{n \to \infty} \ln n = \infty.\]This limit is not useful, but we know that \(\frac{\ln n}{n^{2}} < \frac{1}{n^{2}}\) for all \(n \geq 3\), and since \(\sum_{n=1}^{\infty} \frac{1}{n^{2}}\) converges, we can apply the comparison test.
05
Conclude Using the Comparison Test
Since \(0 \leq \frac{\ln n}{n^{2}} < \frac{1}{n^{2}}\) for \(n \geq 3\) and \(\sum_{n=1}^{\infty} \frac{1}{n^{2}}\) converges, by the Comparison Test, the given series \(\sum_{n=1}^{\infty} \frac{\ln n}{n^{2}}\) also converges.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Convergence of Series
In mathematics, understanding whether a series converges or diverges is incredibly important. A series is essentially the sum of an infinite sequence of numbers. When we say a series converges, it means the sum approaches a fixed value as you keep adding infinitely many terms. If it diverges, the sum does not settle to a value. For instance, the series \( \sum_{n=1}^{\infty} \frac{1}{n} \) diverges, while \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) converges. Learning how to determine convergence helps in various applications, such as calculus and analysis. There are many tests to determine if a series converges, and learning these can greatly enhance your problem-solving skills in math. The big goal is to figure out whether adding up infinitely many numbers ends up being a finite number.
P-Series Test
The P-Series Test is a useful tool for determining the convergence of specific types of series. A p-series is defined as \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), where \( p \) is a positive constant. The behavior of a p-series is heavily dependent on the value of \( p \).
- If \( p > 1 \), the series converges.
- If \( p \leq 1 \), the series diverges.
Logarithmic Functions
Logarithmic functions grow very slowly compared to polynomial or exponential functions. The common form is \( \ln n \), which is the natural logarithm. When evaluating series, understanding this slow growth is key. Logarithmic terms in a series usually serve to compare against other terms, often making the series's behavior less straightforward. Since \( \ln n \) grows slower than \( n^k \) for any positive \( k \), in a setting like \( \frac{\ln n}{n^2} \), the denominator's power will generally dominate, potentially leading to convergence. Understanding these dynamics helps in comparing and determining the nature of series.
Limit Comparison Test
The Limit Comparison Test is a powerful technique used to determine the convergence of a series by comparing it with another series whose convergence is known. Here’s how it works:Consider two series, \( \sum a_n \) and \( \sum b_n \). Compute the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} \). If this limit \( L \) is a positive finite number, then both series \( \sum a_n \) and \( \sum b_n \) will either both converge or both diverge.This test is particularly handy when directly applying the Comparison Test might not be easy. For example, in \( \sum_{n=1}^{\infty} \frac{\ln n}{n^{2}} \), by comparing \( \frac{\ln n}{n^{2}} \) to \( \frac{1}{n^{2}} \), it's seen that even though individual term limits can be tricky or unhelpful, surrounding logic and relationships still ascertain convergence by the Comparison Test.