Problem 29 - Thomas Calculus

Question

Which of the series in Exercises 13 46 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.) $$ \sum_{n=2}^{\infty} \frac{\sqrt{n}}{\ln n} $$

Step-by-Step Solution

Answer

The series \( \sum_{n=2}^{\infty} \frac{\sqrt{n}}{\ln n} \) diverges by the Comparison Test.

Step 1: Identify Type of Series

The given series is \( \sum_{n=2}^{\infty} \frac{\sqrt{n}}{\ln n} \), which is a type of series involving a ratio of functions of \( n \). We need to determine whether it converges or diverges.

Step 2: Apply the Comparison Test

To use the Comparison Test, compare the terms of the given series \( a_n = \frac{\sqrt{n}}{\ln n} \) with a known divergent series. Consider \( b_n = \frac{\sqrt{n}}{n} \), since \( n \) grows faster than \( \ln n \), making \( \ln n < n \). Therefore, \( \frac{1}{\ln n} > \frac{1}{n} \), which implies \( a_n > b_n \). We know from the p-series test that \( \sum \frac{1}{n^k} \) diverges for \( k \leq 1 \), specifically \( b_n = \frac{1}{n^{1/2}} \) is divergent.

Step 3: Conclude Using Comparison

Since \( a_n > b_n \) for large \( n \) and \( \sum b_n \) is a known divergent p-series with \( p=1/2 \), by the Comparison Test, \( \sum a_n \) must also diverge. Thus, \( \sum_{n=2}^{\infty} \frac{\sqrt{n}}{\ln n} \) does not converge.

Key Concepts

Comparison Test

The Comparison Test is a helpful method to determine the convergence or divergence of a series by comparing it to another series with known behavior. This technique involves comparing the given series' terms with those of a series that is already understood to either converge or diverge. To do this, we look for a simpler series, say \( b_n \), such that \( a_n \approx b_n \) for all sufficiently large \( n \).
The Comparison Test works as follows:

• If \( 0 \leq a_n \leq b_n \) and the series \( \sum b_n \) converges, then \( \sum a_n \) also converges.
• If \( 0 \leq b_n \leq a_n \) and the series \( \sum b_n \) diverges, then \( \sum a_n \) also diverges.

In the original exercise, the series \( \sum a_n = \sum_{n=2}^{\infty} \frac{\sqrt{n}}{\ln n} \) was compared to a divergent series \( \sum b_n = \sum_{n=2}^{\infty} \frac{\sqrt{n}}{n} \). As shown, because \( a_n > b_n \) for large enough \( n \) and \( \sum b_n \) diverges, the series \( \sum a_n \) must also diverge.

Divergence

Divergence refers to the behavior of a series where the sum does not approach a finite limit as more terms are added. Simply put, a series that diverges will continue to grow indefinitely or oscillate without settling down to a final value. Determining whether a series converges or diverges is crucial in many areas of mathematics and applied sciences.
In the context of the exercise, we explored the divergence of the series \( \sum_{n=2}^{\infty} \frac{\sqrt{n}}{\ln n} \) by using the Comparison Test. We found that this series diverges since its terms are greater than those of a known divergent series for large \( n \). Knowing the divergence of a series can help predict the behavior of systems in calculus, physics, engineering, and more by indicating instability or unbounded growth.

p-series

A p-series is a special type of series of the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \). The convergence of p-series depends on the value of \( p \). Specifically, a p-series converges if \( p > 1 \) and diverges if \( p \leq 1 \). This simple rule makes p-series a useful reference in evaluating more complex series.
For example, in the original solution, the series \( \sum_{n=2}^{\infty} \frac{1}{n^{1/2}} \) was identified as a divergent p-series since \( p = 1/2 \leq 1 \). Recognizing such relationships allows for quick assessments of other, more complex series by comparison. Thus, the p-series can serve as a benchmark for determining the behavior of other series in mathematical analysis.

Ratio of Functions

The series in the exercise involves expressions like \( \frac{\sqrt{n}}{\ln n} \), showcasing a ratio of functions. Understanding how functions behave in relation to each other, such as which function grows faster, can reveal important information about the series itself.
In our exercise, this understanding was crucial. The function \( n \) in \( \frac{\sqrt{n}}{n} \) grows faster than \( \ln n \) since logarithmic functions grow slower than polynomial functions. By analyzing this ratio, we concluded that \( \frac{\sqrt{n}}{\ln n} \) has larger term sizes than \( \frac{\sqrt{n}}{n} \) as \( n \) becomes large. This insight, combined with other tests like the Comparison Test, helps assess whether a series will converge or diverge. Understanding how these functions interact is a powerful tool in mathematical analysis and helps clarify the solutions to problems involving series.