Question

In the following exercises, use an appropriate test to determine whether the series converges. $$ \left.a_{k}=\left(\frac{k}{k+\ln k}\right)^{2 k} \text { (Hint: } a_{k}=\left(1+\frac{\ln k}{k}\right)^{-(k / \ln k) \ln k^{2}} \cdot\right) $$

Short answer

The series diverges by the Test for Divergence.

Step-by-step solution

Rewrite the Series

The series is given by \( a_k = \left( \frac{k}{k+\ln k} \right)^{2k} \). We begin by rewriting this expression using the hint provided: \( a_k = \left( 1 + \frac{\ln k}{k} \right)^{-\left( \frac{k}{\ln k} \right) \cdot \ln k^2} \).

Recognize it as an Exponential Series

By the expression \( a_k = \left( 1 + \frac{\ln k}{k} \right)^{-\left( \frac{k}{\ln k} \right) \cdot \ln k^2} \), we notice it resembles the exponential form \( (1 + x)^n \). Applying the limit definition of the exponential function, it simplifies to \( e^{-n \cdot x} \).

Simplify the Expression

For large \( k \), the expression becomes: \( a_k \approx \exp\left( - \frac{k}{\ln k} \cdot \ln k^2 \cdot \frac{\ln k}{k} \right) = \exp(-2) \). Thus, each term in the sequence tends to a constant as \( k \to \infty \).

Use the Test for Divergence

The Test for Divergence states that if \( \lim_{k \to \infty} a_k eq 0 \), then the series \( \sum a_k \) diverges. Since \( \lim_{k \to \infty} a_k = e^{-2} eq 0 \), the series does not converge to zero.

Key concepts

Exponential Series

Exponential series are a powerful concept in calculus and analysis, often used to understand complex expressions. Here, we have the series \( a_k = \left( 1 + \frac{\ln k}{k} \right)^{-\left( \frac{k}{\ln k} \right) \cdot \ln k^2} \). This form can be associated with the exponential series because it closely resembles the pattern \((1 + x)^n\), which connects naturally to exponential functions.

The magic of exponential series lies in their relationship with the number \( e \), the base of natural logarithms. The formula \((1 + \frac{x}{n})^n \to e^x\) as \( n \to \infty \), is a fundamental property that helps simplify and solve such problems.

In our series, this relationship explains why simplification leads us to the expression \( e^{-n \cdot x} \). This transformation is crucial in determining the behavior of the series as \( k \) becomes very large.

Test for Divergence

The Test for Divergence is a straightforward yet essential tool in determining the convergence of a series. It states that if the limit of the terms of a series \( \lim_{k \to \infty} a_k eq 0 \), then the series \( \sum a_k \) must diverge. This means that the series does not reach a finite sum and effectively grows indefinitely.

For the given series, \( a_k \approx \exp(-2) \) as \( k \to \infty \). Because \( \exp(-2) eq 0 \), this directly applies the Test for Divergence indicating that the series cannot converge.

Using this test is simple: calculate the limit of the series terms, and if it's anything other than zero, the series diverges. This foundational approach helps quickly identify series that don't sum to a finite value.

Limit Definition

Understanding the limit definition is vital in series analysis. When evaluating whether a series converges, the behavior of the terms as \( k \to \infty \) is crucial. In essence, if the terms of the series do not approach zero, the series can't converge.

In our exercise, the limit \( \lim_{k \to \infty} a_k \) led us to examine the form of \( a_k \) at large \( k \). Using limits, we derived \( a_k \approx e^{-2} \), which shows constancy rather than diminishing towards zero.

The limit definition clarifies how terms that settle on a non-zero constant result in a series that diverges confidently. Grasping this abstract yet practical tool is indispensable for evaluating infinite series effectively.

Expression Simplification

Expression simplification is a crucial skill that makes solving complex series manageable. Starting with the original series expression \( a_k = \left( \frac{k}{k+\ln k} \right)^{2k} \), we rely on algebraic manipulation and hint provided to transform it into a recognizable exponential form.

Switching between forms, particularly into \( a_k = \left( 1 + \frac{\ln k}{k} \right)^{-\left( \frac{k}{\ln k} \right) \cdot \ln k^2} \), allows us to utilize known mathematical principles effectively. Here, it transformed into a situation ripe for applying limits.

By simplifying the sequence down to \( \exp(-2) \), expression simplification aids in seeing clearly whether other series tests, like the Test for Divergence, can be applied. This process underscores the importance of transformation techniques to assess series behavior.