https://file.aiquickdraw.com/mathbot/file/172871581629186xqa7v0.jpg

We are tasked with determining the convergence or divergence of the series:

\[
\sum_{n=1}^{\infty} \frac{1 - \cos^2 n}{n^{3/2}}
\]

using the **Direct Comparison Test**. 

### Step 1: Simplify the expression

First, simplify the term \( 1 - \cos^2 n \). Using the trigonometric identity:
\[
1 - \cos^2 n = \sin^2 n
\]
So, the series becomes:
\[
\sum_{n=1}^{\infty} \frac{\sin^2 n}{n^{3/2}}
\]

### Step 2: Establish a comparison function

To use the **Direct Comparison Test**, we need to find a function \( b_n \) such that:
\[
0 \leq a_n \leq b_n
\]
for all \( n \), where \( a_n = \frac{\sin^2 n}{n^{3/2}} \). We need a function \( b_n \) whose convergence or divergence is easier to determine.

We know that \( \sin^2 n \) is bounded:
\[
0 \leq \sin^2 n \leq 1
\]
Thus, we can compare:
\[
0 \leq \frac{\sin^2 n}{n^{3/2}} \leq \frac{1}{n^{3/2}}
\]

### Step 3: Analyze the comparison series

The series we are comparing with is:
\[
\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}
\]
This is a **p-series** with \( p = 3/2 \). The p-series \( \sum \frac{1}{n^p} \) converges if \( p > 1 \). Since \( \frac{3}{2} > 1 \), the comparison series converges.

### Step 4: Apply the Direct Comparison Test

Since:
\[
0 \leq \frac{\sin^2 n}{n^{3/2}} \leq \frac{1}{n^{3/2}}
\]
and \( \sum \frac{1}{n^{3/2}} \) converges, by the **Direct Comparison Test**, the original series:
\[
\sum_{n=1}^{\infty} \frac{\sin^2 n}{n^{3/2}}
\]
**converges** as well.

---

Would you like any further details or have any questions on this explanation?

Here are five related questions you could explore:

1. What is the **Limit Comparison Test** and how does it differ from the Direct Comparison Test?
2. Can the Direct Comparison Test be applied to divergent series, and under what conditions?
3. How would the analysis change if the exponent on \( n \) were different, say \( n^{5/2} \) instead of \( n^{3/2} \)?
4. Can you apply the **Integral Test** to verify the convergence of the series \( \sum \frac{1}{n^{3/2}} \)?
5. How does the behavior of \( \sin^2 n \) affect the overall convergence if the power of \( n \) in the denominator were smaller?

**Tip**: When dealing with trigonometric functions in series, always check their bounds to simplify comparisons for convergence tests.

Use the direct comparison test to determine if the series ∑ (1 - cos²(n)) / n^(3/2) converges or diverges.

Learn how to determine if the series involving (1 - cos²(n)) / n^(3/2) converges or diverges using the Direct Comparison Test. This problem explores the simplification of trigonometric terms and comparison to a p-series.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation