Problem 42 - 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.) $$ \sum_{n=3}^{\infty} \frac{7}{\sqrt{n+1} \ln \sqrt{n+1}} $$

Step-by-Step Solution

Answer

The series diverges.

Step 1: Identify the Series

We are given the series \( \sum_{n=3}^{\infty} \frac{7}{\sqrt{n+1} \ln \sqrt{n+1}} \). Our task is to determine whether this series converges or diverges.

Step 2: Simplify the Terms

First, let's simplify the expression inside the series. Rewriting the denominator as \( \sqrt{n+1} \ln \sqrt{n+1} = (n+1)^{1/2} \cdot \ln((n+1)^{1/2}) \), we have the series \( \sum_{n=3}^{\infty} \frac{7}{(n+1)^{1/2} \cdot \frac{1}{2} \ln(n+1)} = \sum_{n=3}^{\infty} \frac{14}{(n+1)^{1/2} \ln(n+1)} \).

Step 3: Test for Convergence using the Integral Test

To apply the integral test, consider the function \( f(x) = \frac{14}{(x+1)^{1/2} \ln(x+1)} \), where \( x \geq 3 \). This function is positive, continuous, and decreasing for \( x \geq 3 \). The integral to test is \( \int_{3}^{\infty} \frac{14}{(x+1)^{1/2} \ln(x+1)} \, dx \).

Step 4: Evaluate the Integral

Let \( u = \ln(x+1) \). Then \( du = \frac{1}{x+1} \, dx \). Thus, \( dx = (x+1) \, du \). Substitute back into the integral to get:\[ \int \frac{14}{(x+1)^{1/2} \cdot u} \cdot (x+1) \, du = \int \frac{14(x+1)^{1/2}}{u} \, du. \]The substitution complicates integration directly, but we notice that since the integral involves \( \ln(x+1) \) in the denominator, the function doesn't converge as \( x \to \infty \).

Step 5: Compare with a Known Divergent Series

To better understand the divergence, compare our series with the p-series \( \sum_{n=3}^{\infty} \frac{1}{n^{1/2}} \), which diverges because its power \( \frac{1}{2} \leq 1 \). Our series has an additional \( \ln(n+1) \) term in the denominator, which only strengthens the divergence.

Step 6: Conclude Based on the Integral Test and Comparison

Both the integral and the comparison test with the known divergent p-series suggest that the series \( \sum_{n=3}^{\infty} \frac{14}{(n+1)^{1/2} \ln(n+1)} \) diverges.

Key Concepts

Integral Test

The integral test is a powerful tool used to determine the convergence or divergence of infinite series. When applying the integral test, we first identify a corresponding function, \( f(x) \), that is positive, continuous, and decreasing over the interval \([N, \infty)\). This function needs to exactly match the terms of the series we are examining. To use the integral test:

• Set up the improper integral \( \int_{N}^{\infty} f(x) \, dx \)
• Evaluate whether this integral converges or diverges

If the integral converges, then the series \( \sum_{n=N}^{\infty} f(n) \) also converges. If the integral diverges, the series diverges as well. In our problem, the series \( \sum_{n=3}^{\infty} \frac{14}{(n+1)^{1/2} \ln(n+1)} \) was tested by setting an improper integral. Despite the more complex evaluation due to the \( \ln(n+1) \) term, the divergence of this integral implies the divergence of the series.

Divergent Series

A divergent series is an infinite series that does not have a finite sum. Determining whether a series diverges is crucial in analysis because it shows that piling up the terms of the series infinitely will add up to infinity or fail to settle at any particular number. For a series \( \sum a_n \), if \( \lim_{n \to \infty} \sum_{i=1}^n a_i \) does not exist or is infinite, then the series diverges. The series in question, \( \sum_{n=3}^{\infty} \frac{14}{(n+1)^{1/2} \ln(n+1)} \), diverges because the improper integral used in the integral test diverged. Additionally, when compared to known divergent series, it becomes clear that the series will also diverge, illustrating how additional terms like logarithms interact in diverging behavior.

P-Series

A p-series is a special type of series of the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), where \( p \) is a constant. The convergence of p-series depends heavily on the value of \( p \):

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

In evaluating the given series, the comparison to a p-series like \( \sum \frac{1}{n^{1/2}} \) (a divergent p-series since \( 1/2 \leq 1 \)) helped establish benchmark behavior. Knowing that a p-series where \( p = \frac{1}{2} \) diverges, it provides evidence that the original series, which is even larger without the log term due to additional complexity, diverges likewise.

Comparison Test

The comparison test is a method used to determine the convergence or divergence of a series by comparing it to another series whose convergence properties are already known. It operates under these principles:

• If \( 0 \leq a_n \leq b_n \) for all \( n \) and \( \sum b_n \) converges, then \( \sum a_n \) also converges.
• If \( a_n \geq b_n \) for all \( n \) and \( \sum b_n \) diverges, then \( \sum a_n \) diverges as well.

Using the comparison test in the problem, we compared \( \sum \frac{14}{(n+1)^{1/2} \ln(n+1)} \) to the divergent series \( \sum \frac{1}{n^{1/2}} \). The presence of an additional logarithm term in the denominator of our series suggests \( a_n < b_n \). This aligns with the divergence of \( \sum \frac{1}{n^{1/2}} \), indicating the original series also diverges, as its behavior would not be less than a known divergent reference.