Question

Use the test of your choice to determine whether the following series converge. $$\sum_{k=2}^{\infty} \frac{1}{k^{\ln k}}$$

Short answer

#Answer# The series $\sum_{k=2}^{\infty} \frac{1}{k^{\ln k}}$ converges.

Step-by-step solution

Identify the series and corresponding function

The series is given as: $$\sum_{k=2}^{\infty} \frac{1}{k^{\ln k}}$$ The corresponding function for the Integral Test is: $$f(k)=\frac{1}{k^{\ln k}}$$ Step 2: Setting up the integral

Set up integral

To apply the Integral Test, we want to evaluate the integral of the corresponding function from \(k=2\) to \(k=\infty\). Thus, we set up the integral as follows: $$\int_{2}^{\infty} \frac{1}{k^{\ln k}} dk$$ Step 3: Solving the integral

Solve the integral

To solve the integral, we will do a substitution. Let \(u = \ln k\). Then, \(du = \frac{1}{k} dk\). So when \(k=2\), \(u=\ln(2)\), and as \(k \to \infty\), \(u \to \infty\). Replacing \(k\) and \(dk\) in the integral, we get: $$\int_{\ln(2)}^{\infty} \frac{1}{e^{u^2}} du$$ Now, let's compare this integral with the integral of Gaussian function \(\int_{\ln(2)}^{\infty} e^{-u^2} du\). Since \(\frac{1}{e^{u^2}} = e^{-u^2}\), we have: $$\int_{\ln(2)}^{\infty} \frac{1}{e^{u^2}} du < \int_{\ln(2)}^{\infty} e^{-u^2} du$$ We know that the integral of Gaussian function converges; therefore, by Comparison Test, our integral also converges. Step 4: Conclusion

Conclude

Since the integral of the function converges, by the Integral Test, the original series converges as well. Thus, we can conclude that the series: $$\sum_{k=2}^{\infty} \frac{1}{k^{\ln k}}$$ converges.

Key concepts

Convergence of Series

Understanding the convergence of a series is an essential topic in calculus. When we speak of convergence, we are looking to see whether the sum of infinite terms in a series approaches a finite number. If it does, we say the series converges. Otherwise, the series diverges. Testing for convergence often requires various methods and tests. One common approach is to apply the Integral Test. This involves translating the series into a continuous function and evaluating the improper integral over the desired limits. If the definite integral yields a finite result, then the series can be concluded to converge. Conversely, if the integral diverges, the series does as well. Understanding these principles helps in analyzing complex series and making sense of their behavior over infinity.

Comparison Test

The Comparison Test is a valuable tool for determining series convergence. It compares a given series to another series whose convergence properties are already known. There are two primary forms: - **Direct Comparison Test:** This involves directly comparing the terms of two series. If each term of your series is smaller than a converging comparison series, then your series also converges. - **Limit Comparison Test:** This form is useful when direct comparison isn't straightforward. By taking the limit of the ratio of the series terms as they approach infinity, we can determine convergence based on known series behavior. In our solution, the Comparison Test was used to compare the transformed integral to a Gaussian function's integral, which is known to converge. This verification finalizes that our integral and thus the original series converges as well.

Gaussian Function

The Gaussian function, characterized by its bell-like curve, plays a crucial role in mathematics and statistics. Its general form is given by the exponential function \( e^{-x^2} \). It is especially important in the context of series convergence because the integral of this function over the entire real line is known to converge. In our step-by-step solution, we observed this function's integral to draw conclusions about the original series' behavior. By known properties, the definite integral \( \int_{a}^{b} e^{-x^2} dx \) converges to a finite number, helping us conclude that similar integrals related to our series will also converge. This real-world utility of the Gaussian function aids significantly in confirming convergence characteristics in mathematical analysis.

Substitution Method in Integrals

The substitution method is a handy technique in calculus for simplifying integrals. Often, when dealing with complex expressions, we replace a part of the integrand with a single variable to transform the expression into a more manageable form. In the exercise at hand, substitution was employed by setting \( u = \ln k \), leading to \( du = \frac{1}{k} dk \). This change of variables turns the integral into a form reminiscent of the Gaussian integral, making the evaluation straightforward. Substitution not only simplifies integration but can also convert potentials of divergence into recognizable convergent forms, like in our exercise. Mastery of this method provides a foundation for solving a wide range of integrated problems effectively.