Problem 39 - Thomas Calculus

Question

In Exercises \(17-46,\) use any method to determine if the series converges or diverges. Give reasons for your answer. $$ \sum_{n=2}^{\infty} \frac{-n}{(\ln n)^{n}} $$

Step-by-Step Solution

Answer

The series converges by the Ratio Test since the limit is less than 1.

Step 1: Understand the Series

The series in question is \[ \sum_{n=2}^{\infty} \frac{-n}{(\ln n)^{n}}. \]We need to determine if this series converges or diverges.

Step 2: Identify the Test

The terms of the series are \[ a_n = \frac{-n}{(\ln n)^{n}}. \]Because of the factorial power of \((\ln n)^n\), we can use the Ratio Test to analyze the convergence.

Step 3: Apply the Ratio Test

For the Ratio Test, we calculate the limit \[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|. \]Substitute \( a_n = \frac{-n}{(\ln n)^{n}} \) and \( a_{n+1} = \frac{-(n+1)}{(\ln(n+1))^{n+1}} \) into the equation.

Step 4: Simplify the Ratio

The ratio \[ \frac{a_{n+1}}{a_n} = \frac{-(n+1)}{(\ln(n+1))^{n+1}} \times \frac{(\ln n)^{n}}{-n} = \frac{(n+1)(\ln n)^n}{n (\ln(n+1))^{n+1}}. \]Let's simplify this expression before taking the limit.

Step 5: Simplify Further

Rearrange and factor terms: \[ \frac{a_{n+1}}{a_n} = \frac{n+1}{n} \times \left( \frac{\ln n}{\ln(n+1)} \right)^n \times \frac{1}{\ln(n+1)}. \] Use the approximation \( \ln(n+1) \approx \ln n \) for large \( n \).

Step 6: Calculate the Limit

For large \( n \), the terms \( \frac{n+1}{n} \to 1 \) and \( \left(\frac{\ln n}{\ln(n+1)}\right)^n \to \frac{1}{e} \) because \( \ln(n+1) \approx \ln n + \frac{1}{n} \) implies \( \left( \frac{\ln n}{\ln(n+1)} \right)^n \to e^{-1} \) and \( \frac{1}{\ln(n+1)} \to 0 \). Therefore:\[ L = 1 \times \frac{1}{e} \times 0 = 0. \]

Step 7: Conclusion Using the Ratio Test

The limit \( L = 0 < 1 \), so by the Ratio Test, the series \( \sum_{n=2}^{\infty} \frac{-n}{(\ln n)^{n}} \) converges.

Key Concepts

Ratio Test

The Ratio Test is a powerful tool in calculus for determining the convergence of an infinite series. It involves taking the limit of the absolute value of the ratio of consecutive terms in a series. For a series \( \sum a_n \), we define the test as follows:- Compute the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).The Ratio Test states that:

• If \( L < 1 \), the series converges absolutely.
• If \( L > 1 \), the series diverges.
• If \( L = 1 \), the test is inconclusive.

The simplicity of the Ratio Test makes it particularly useful for series with factorial or exponential terms. In our exercise, the term is \( a_n = \frac{-n}{(\ln n)^n} \), which presents a natural case to apply this test due to the exponential appearance of \( (\ln n)^n \). It's important to carefully simplify the ratio before evaluating the limit to maximize test accuracy.

Infinite Series

An infinite series is the sum of an infinite sequence of terms. Written in the form \( \sum_{n=1}^{\infty} a_n \), it extends indefinitely but follows a particular formula or pattern for its terms \( a_n \).Infinite series can be:

• Convergent: The sum approaches a finite limit as more terms are added.
• Divergent: The sum does not approach a finite limit.

To determine the nature of a series, mathematicians use various convergence tests, such as the Ratio Test, Comparison Test, or the Integral Test. Understanding the behavior of series is crucial in many areas of mathematics, such as analysis and differential equations. Convergence and divergence are bigger concepts that impact how functions behave over infinite intervals.

Mathematical Convergence

Mathematical convergence refers to the idea that a sequence or series approaches a specific value as the number of terms increases indefinitely. For a series \( \sum_{n=1}^{\infty} a_n \) to be convergent, the partial sums \( S_N = \sum_{n=1}^{N} a_n \) must approach a finite value \( S \) as \( N \to \infty \).In practical terms:

• A convergent series sums to a number you can pinpoint with enough terms.
• A divergent series grows without bound or oscillates without settling down.

Convergence is fundamental in analysis because it determines if operations on infinite series yield useful results. For example, convergence allows for meaningful integration and differentiation of function series and ensures stability in solutions of differential equations.

Logarithmic Functions

Logarithmic functions, represented as \( \log_b(x) \) where \( b \) is the base, are the inverse operations of exponentiation. In this exercise, we encounter the natural logarithm \( \ln(x) \), which uses the mathematical constant \( e \) (approximately 2.718) as its base.Key properties of logarithmic functions include:

• The function \( \ln(x) \) is only defined for \( x > 0 \).
• \( \ln(1) = 0 \) because \( e^0 = 1 \).
• \( \ln(xy) = \ln(x) + \ln(y) \), known as the product rule.
• \( \ln\left(\frac{x}{y}\right) = \ln(x) - \ln(y) \), the quotient rule.

In the context of convergence, understanding logarithms is crucial because the terms of our series \( a_n = \frac{-n}{(\ln n)^n} \) involve various logarithmic transformations. The behavior of logarithmic functions dramatically influences the convergence of such series through their growth rates and asymptotic properties.