Problem 22 - 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.) $$ \sum_{n=2}^{\infty} \frac{\ln n}{\sqrt{n}} $$

Step-by-Step Solution

Answer

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

Step 1: Analyze the terms of the series

The given series is \( \sum_{n=2}^{\infty} \frac{\ln n}{\sqrt{n}} \). To analyze this series, we observe the general term \( a_n = \frac{\ln n}{\sqrt{n}} \). Notably, as \( n \) increases, \( \ln n \) grows, but it grows slower than any polynomial power of \( n \).

Step 2: Apply the Limit Comparison Test with a known series

Let's compare \( a_n = \frac{\ln n}{\sqrt{n}} \) with the series \( b_n = \frac{1}{n^{3/4}} \), which is a \( p \)-series with \( p = 3/4 \). First, note that \( \int \frac{1}{n^{3/4}} \, dx \) diverges as \( p = 3/4 < 1 \).

Step 3: Evaluate the Limit Comparison Ratio

Recall the Limit Comparison Test: determine \( L = \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{\ln n}{\sqrt{n}}}{\frac{1}{n^{3/4}}} = \lim_{n \to \infty} \ln n \cdot n^{1/4} \). Notice here that for large \( n \), \( n^{1/4} \) grows faster than \( \ln n \), suggesting \( L = \infty \), indicating the original series behaves like a divergent \( p \)-series as \( b_n \).

Step 4: Conclusion on convergence

Since \( L = \infty \) in the Limit Comparison Test and \( \sum_{n=2}^{\infty} b_n \) with \( b_n = \frac{1}{n^{3/4}} \) diverges, the series \( \sum_{n=2}^{\infty} \frac{\ln n}{\sqrt{n}} \) also diverges due to \( n^{1/4} \) dominating the growth of \( \ln n \).

Key Concepts

Limit Comparison Test

The Limit Comparison Test is a useful tool when testing the convergence or divergence of a series that appears complicated or isn't a standard form. It relies on comparing the series under consideration to another series with known convergence behavior. Consider two series, \( \sum a_n \) and \( \sum b_n \). If \( a_n \) and \( b_n \) are positive for sufficiently large \( n \), and \( \lim_{n \to \infty} \frac{a_n}{b_n} = L \) for some finite \( L > 0 \), then both series will either converge or diverge together.
For the original problem, we compare \( a_n = \frac{\ln n}{\sqrt{n}} \) with \( b_n = \frac{1}{n^{3/4}} \) because understanding a simple \( p \)-series helps deduce the behavior of the complex original series.

P-Series

A \( p \)-series is a series of the form \( \sum \frac{1}{n^p} \), which converges if \( p > 1 \) and diverges if \( p \leq 1 \). Knowing the behavior of \( p \)-series is crucial for comparison tests since they serve as reference points.In the original example, we used \( b_n = \frac{1}{n^{3/4}} \), a \( p \)-series with \( p = 3/4 \). Given that \( p < 1 \), the series diverges. This hints about the original series' behavior when compared using the Limit Comparison Test.
Bridging familiar \( p \)-series with given series helps in predicting whether the complex series converges or diverges.

Divergence

Divergence in series occurs when the sum of its terms increases indefinitely as more terms are added. Knowing if a series diverges or not is essential when analyzing its behavior.For instance, in applying the Limit Comparison Test in the given exercise, we found that both the original series \( \sum \frac{\ln n}{\sqrt{n}} \) and its comparison series \( \sum \frac{1}{n^{3/4}} \) diverge. This conclusion is based on the fact that the sequences involved have terms that grow extensively as \( n \rightarrow \infty \), leading to a series that does not settle down to a finite sum.
Understanding divergence helps in pointing out whether the series sums are finite or stretch endlessly.

Ln Function Growth

The \( \ln n \) function, or the natural logarithm, grows as \( n \) increases, but its growth is slower than polynomial functions. This growth property is key when considering series like \( \sum \frac{\ln n}{\sqrt{n}} \).In the example, while \( \ln n \) grows, the comparison was made with \( n^{1/4} \), which increases faster. To simplify why, consider the functions: a logarithmic function like \( \ln n \) grows slower in comparison to any algebraic expression of \( n \) with a positive exponent.
Recognizing this growth rate allows us to understand how different terms affect a series' tendency to diverge based on their rate of increase.