Problem 20 - Thomas Calculus

Question

Which of the series in Exercises \(11-40\) converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series' convergence or divergence.) $$ \sum_{n=2}^{\infty} \frac{\ln n}{\sqrt{n}} $$

Step-by-Step Solution

Answer

The series \( \sum_{n=2}^{\infty} \frac{\ln n}{\sqrt{n}} \) diverges since its terms do not approach 0 as \( n \to \infty \).

Step 1: Identify the Series

We are given the series \( \sum_{n=2}^{\infty} \frac{\ln n}{\sqrt{n}} \). The task is to determine whether this series converges or diverges.

Step 2: Test for Convergence Using Limit Comparison

Let's use the Limit Comparison Test. For comparison, consider the series \( \sum_{n=2}^{\infty} \frac{1}{n^{3/2}} \), which is a p-series with \( p = \frac{3}{2} \) and therefore convergent. Now, compute the limit \( L \): \[L = \lim_{n \to \infty} \frac{\frac{\ln n}{\sqrt{n}}}{\frac{1}{n^{3/2}}} = \lim_{n \to \infty} \frac{\ln n \cdot n^{3/2}}{\sqrt{n}} = \lim_{n \to \infty} \ln n \cdot n = \infty\]

Step 3: Interpret the Limit

The limit \( L \) calculated in Step 2 is infinite. The Limit Comparison Test requires a finite positive limit to conclude convergence; since \( L = \infty \), this approach doesn't show convergence in relation to our comparison series.

Step 4: Alternative Test for Divergence

Let's use the Divergence Test, which states that if the terms \( a_n \) do not approach 0, the series \( \sum a_n \) diverges. For our series, \( a_n = \frac{\ln n}{\sqrt{n}} \). Clearly, as \( n \to \infty \), \( a_n \to \infty \), instead of approaching 0. Therefore, the series diverges.

Key Concepts

Limit Comparison Test

The Limit Comparison Test is an efficient method to determine if a series converges or diverges by comparing it with another series whose behavior is known. This test is particularly useful when dealing with series that are difficult to evaluate directly.

To apply the Limit Comparison Test, follow these steps:

• Choose a suitable comparison series, typically one that you know converges or diverges, such as a p-series.
• Calculate the limit \( L \) of the ratio of the terms of the two series as \( n \) approaches infinity. This ratio is between the general term of the given series and the term of the chosen comparison series.
• The series you are examining behaves similarly to the comparison series if the limit \( L \) is a positive, finite number.

In the step-by-step solution provided, the Limit Comparison Test involved the series \( \sum \frac{\ln n}{\sqrt{n}} \) and the p-series \( \sum \frac{1}{n^{3/2}} \). Calculating the limit resulted in infinity, indicating that the test was inconclusive for this comparison. This led to trying another approach such as the Divergence Test.

Divergence Test

The Divergence Test is one of the simplest tools for identifying series that diverge. It's based on a straightforward idea: if the terms of a series do not approach zero, then the series cannot possibly converge. This test is often the first check performed on any series.

The steps to apply the Divergence Test include:

• Identify the general term \( a_n \) of your series.
• Check the limit of \( a_n \) as \( n \to \infty \). If \( \lim_{n \to \infty} a_n eq 0 \), then \( \sum a_n \) diverges.

In the provided solution, the given series \( \sum \frac{\ln n}{\sqrt{n}} \) had terms that did not tend toward zero as \( n \to \infty \), making it clear that the series diverges. Therefore, the Divergence Test was able to confirm divergence when the Limit Comparison Test could not.

P-Series Test

P-Series are a family of series that take the form \( \sum \frac{1}{n^p} \). Their convergence or divergence depends heavily on the value of \( p \):

• If \( p > 1 \), the p-series converges.
• If \( p \leq 1 \), the p-series diverges.

Understanding p-series is critical when using the Limit Comparison Test, as it provides standard cases for comparison. In this exercise, a p-series \( \sum \frac{1}{n^{3/2}} \) with \( p = \frac{3}{2} > 1 \) was utilized. This meant that the comparison series converged, serving as a potential reference for the original series. However, in this scenario, despite the p-series converging, the original series did not, which was clear from applying the Limit Comparison Test and resulting limit being infinite. Thus, it reinforced the need to utilize multiple tests to determine convergence or divergence.