Chapter 4: Problem 22
Use the root test to determine whether \(\sum_{m=1}^{\infty} a_{n}\) converges, where \(a_{n}\) is as follows. $$ a_{n}=\frac{(\ln n)^{2 n}}{n^{n}} $$
Short Answer
Expert verified
The series \( \sum_{n=1}^{\infty} a_n \) converges by the root test.
Step by step solution
01
Understand the Series
We have a series represented by \( \sum_{n=1}^{\infty} a_n \) where \( a_n = \frac{(\ln n)^{2n}}{n^n} \). To determine its convergence, we will apply the root test.
02
Apply the Root Test Formula
For the root test, we consider \( \lim_{n \to \infty} \sqrt[n]{|a_n|} \). Here, \( a_n = \frac{(\ln n)^{2n}}{n^n} \), so our next step is to find \( \sqrt[n]{|a_n|} = \sqrt[n]{\frac{(\ln n)^{2n}}{n^n}} \).
03
Simplify the Root Expression
Using properties of exponents, \( \sqrt[n]{\frac{(\ln n)^{2n}}{n^n}} = \frac{(\ln n)^2}{n} \). This is because \( \sqrt[n]{x^n} = x \). Therefore, the expression simplifies to \( \frac{(\ln n)^2}{n} \).
04
Evaluate the Limit
We need to find \( \lim_{n \to \infty} \frac{(\ln n)^2}{n} \). Note that as \( n \to \infty \), \( \ln n \to \infty \) as well, but \( (\ln n)^2 \) grows slower than \( n \). Therefore, \( \lim_{n \to \infty} \frac{(\ln n)^2}{n} = 0 \).
05
Conclusion Using the Root Test
The root test states that if \( \lim_{n \to \infty} \sqrt[n]{|a_n|} < 1 \), then the series converges. Since we found that \( \lim_{n \to \infty} \frac{(\ln n)^2}{n} = 0 \), which is less than 1, the series \( \sum_{n=1}^{\infty} a_n \) converges.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Series Convergence
In calculus, when we discuss series convergence, we are trying to determine whether a given infinite series has a sum that approaches a finite number. An infinite series is expressed as the sum of terms in the form \( \sum_{n=1}^{\infty} a_n \). The concept of convergence is crucial in understanding the behavior of series as it tells us whether adding infinitely many terms results in a finite outcome. If the series converges, we say the sum is meaningful and equals a specific value. If it diverges, the sum is infinite or does not settle at a fixed value.
Several tests are available to analyze convergence, such as the ratio test, comparison test, and root test. In our problem, we use the root test because it is particularly useful when series terms include exponential expressions, like \( a_n = \frac {(\ln n)^{2n}}{n^n} \), as it can help efficiently handle such components.
Several tests are available to analyze convergence, such as the ratio test, comparison test, and root test. In our problem, we use the root test because it is particularly useful when series terms include exponential expressions, like \( a_n = \frac {(\ln n)^{2n}}{n^n} \), as it can help efficiently handle such components.
Root Test
The root test is a powerful tool in calculus for determining the convergence of an infinite series. It is especially helpful when dealing with terms raised to the power of \( n \), as it leverages the properties of roots and limits. Given a series \( \sum_{n=1}^{\infty} a_n \), the root test states that we need to consider the limit \( \lim_{n \to \infty} \sqrt[n]{|a_n|} \).
We look at the behavior of this expression as \( n \) approaches infinity:
For example, in our problem, the series is represented by \( a_n = \frac{(\ln n)^{2n}}{n^n} \), and simplification leads us to analyze \( \frac{(\ln n)^2}{n} \). When we see that this limit equals 0—less than 1—the root test guarantees that the series converges.
We look at the behavior of this expression as \( n \) approaches infinity:
- If the limit is less than 1, the series converges.
- If the limit is greater than 1, the series diverges.
- If the limit equals 1, the test is inconclusive.
For example, in our problem, the series is represented by \( a_n = \frac{(\ln n)^{2n}}{n^n} \), and simplification leads us to analyze \( \frac{(\ln n)^2}{n} \). When we see that this limit equals 0—less than 1—the root test guarantees that the series converges.
Natural Logarithm
The natural logarithm is denoted as \( \ln(n) \) and is the logarithm to the base \( e \), where \( e \approx 2.71828 \). In mathematics, the natural logarithm is an essential function, especially in calculus and analysis.
It arises in various scenarios because it simplifies the differentiation and integration of exponential functions. In this exercise, the natural logarithm plays a crucial role in the expression \( a_n = \frac{(\ln n)^{2n}}{n^n} \). As \( n \) increases, \( \ln n \) also increases, but at a much slower rate compared to \( n \). This characteristic is vital in determining the convergence of the series when applying the root test.
The complexity of the logarithmic growth rate is why the square of \( \ln n \), i.e., \( (\ln n)^2 \), grows slower than linear functions like \( n \). This makes \( \ln n \) a delicate yet critical factor in evaluating expressions that involve such terms.
It arises in various scenarios because it simplifies the differentiation and integration of exponential functions. In this exercise, the natural logarithm plays a crucial role in the expression \( a_n = \frac{(\ln n)^{2n}}{n^n} \). As \( n \) increases, \( \ln n \) also increases, but at a much slower rate compared to \( n \). This characteristic is vital in determining the convergence of the series when applying the root test.
The complexity of the logarithmic growth rate is why the square of \( \ln n \), i.e., \( (\ln n)^2 \), grows slower than linear functions like \( n \). This makes \( \ln n \) a delicate yet critical factor in evaluating expressions that involve such terms.
Infinite Series
An infinite series in calculus is a sum of an endless sequence of terms, expressed in the form \( \sum_{n=1}^{\infty} a_n \). Infinite series can either converge or diverge, and understanding their behavior is fundamental in calculus, paving the way to explore even more advanced topics such as Fourier series or series solutions to differential equations.
To determine a series's convergence status, one must analyze the accumulated effect of adding an infinite number of terms. Concepts like the root test come into play, facilitating this analysis by providing systematic ways to gauge the growth or decay of series terms.
In our example, the series \( \sum_{m=1}^{\infty} \frac{(\ln n)^{2n}}{n^n} \) consists of terms involving exponential expressions. We took advantage of the root test to conclude that this particular series converges. Recognizing whether a series converges can imply the presence of a meaningful sum—a point of interest in both theoretical and practical applications.
To determine a series's convergence status, one must analyze the accumulated effect of adding an infinite number of terms. Concepts like the root test come into play, facilitating this analysis by providing systematic ways to gauge the growth or decay of series terms.
In our example, the series \( \sum_{m=1}^{\infty} \frac{(\ln n)^{2n}}{n^n} \) consists of terms involving exponential expressions. We took advantage of the root test to conclude that this particular series converges. Recognizing whether a series converges can imply the presence of a meaningful sum—a point of interest in both theoretical and practical applications.
