Question

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

Short answer

Answer: The series \(\sum_{k=2}^{\infty} \frac{1}{k^2 \ln k}\) converges.

Step-by-step solution

Identify the Comparison Series

We will compare the given series to a p-series with \(p = 2\). This series converges, as \(p > 1\). The comparison series is given as: $$\sum_{k=2}^{x} \frac{1}{k^{2}}$$

Perform the Comparison Test

We now perform the Comparison Test by comparing the terms of the given series and the p-series: $$\frac{1}{k^2 \ln k} \leq \frac{1}{k^2}$$ To determine if the inequality holds true for all \(k \geq 2\), we can divide both sides by \(\frac{1}{k^2}\): $$\ln k \geq 1$$ Considering that the natural logarithm function is strictly increasing, and since \(\ln(2) > 0\), we can conclude that the inequality holds for all \(k \geq 2\). Therefore, the terms of the given series are, in fact, smaller than or equal to the corresponding terms of the convergent p-series.

Determine Convergence

Since the terms of the given series are smaller than or equal to the terms of a convergent p-series, we can conclude that the given series also converges. This is based on the Comparison Test. Thus, the series: $$\sum_{k=2}^{x} \frac{1}{k^{2} \ln k}$$ converges.

Key concepts

p-series

A p-series is a specific type of infinite series that looks like this: \[ \sum_{k=1}^{\infty} \frac{1}{k^{p}} \] where \(p\) is a positive constant. Understanding the behavior of p-series is crucial because it helps determine if the series converges or diverges.

• Convergence of a p-series: A p-series will converge if \(p > 1\). This is because the sum approaches a finite limit as iterations of terms repeat infinitely.
• Divergence of a p-series: If \(p \leq 1\), the p-series will diverge, meaning its sum will continue to increase indefinitely without nearing a specific value.

In the original exercise, the series \(\sum_{k=2}^{x} \frac{1}{k^{2}}\) is a converging p-series with \(p = 2\). Recognizing and comparing to such known forms can simplify the process of convergence testing for more complex series.

Convergence

Convergence is a central concept when working with series. It refers to whether the sum of a series approaches a distinct limit as the number of terms increases toward infinity.

• Convergent Series: If the partial sums \(S_n = a_1 + a_2 + a_3 + \ldots + a_n\) approach a finite number as \(n\) increases, the series converges.
• Divergent Series: If these partial sums do not approach a specific number or increase indefinitely, the series is said to diverge.

Testing for convergence often involves using comparison with known, simpler series, such as the p-series or geometric series. In our exercise, applying the Comparison Test to the given series helped identify its convergence status through comparing it with the well-known convergent \(\sum_{k=2}^{x} \frac{1}{k^{2}}\) series.

Logarithm Function

The logarithm function, especially the natural logarithm denoted as \(\ln(x)\), plays a critical role in many mathematical contexts, such as growth rates and time-based events. In this exercise, it appears in the term \(\ln(k)\) of the series.

• Properties of the Logarithm Function:
  • The logarithm function grows slowly compared to polynomial functions. For example, as \(k\) increases, \(\ln(k)\) increases but at a decreasing rate when compared to \(k^2\).
  • The logarithm function is strictly increasing, meaning if \(a < b\), then \(\ln(a) < \ln(b)\).
• Logarithms in Convergence:Being an increasing function, knowing that \(\ln(k) \geq 1\) for \(k \geq 2\) aids our understanding of how terms compare in magnitude with other series elements.

In our series, understanding these properties was pivotal in confirming that the inequality \(\ln(k) \geq 1\) holds true for all terms, supporting the final claim of convergence through the Comparison Test.