https://file.aiquickdraw.com/mathbot/file/17300509030425ukchu3s.jpg

To solve this problem using the Limit Comparison Test, let’s proceed step-by-step:

1. **Identify the Given Series**:  
   The series provided is:
   \[
   \sum_{n=2}^{\infty} a_n = \sum_{n=2}^{\infty} \frac{16n - 5}{4n\sqrt{n}}
   \]

2. **Simplify \(a_n\)**:
   Simplify the term \( a_n = \frac{16n - 5}{4n\sqrt{n}} \):
   \[
   a_n = \frac{16n - 5}{4n\sqrt{n}} = \frac{16n}{4n\sqrt{n}} - \frac{5}{4n\sqrt{n}} = \frac{4}{\sqrt{n}} - \frac{5}{4n\sqrt{n}}
   \]
   As \(n \to \infty\), the term \(\frac{5}{4n\sqrt{n}}\) becomes negligible, so \( a_n \approx \frac{4}{\sqrt{n}} \).

3. **Choose a Comparison Series**:  
   Based on the form of \(a_n\), we can select a comparison series:
   \[
   b_n = \frac{c}{n^{p}}
   \]
   with \( c = 4 \) and \( p = \frac{1}{2} \), so:
   \[
   b_n = \frac{4}{n^{1/2}}
   \]

4. **Limit Comparison Test**:  
   To use the Limit Comparison Test, calculate:
   \[
   \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{4}{\sqrt{n}}}{\frac{4}{\sqrt{n}}} = 1
   \]
   Since this limit is a finite, positive number (1), the Limit Comparison Test applies.

5. **Determine Convergence of \( b_n \)**:  
   The series \( \sum b_n = \sum \frac{4}{n^{1/2}} \) is a \(p\)-series with \(p = \frac{1}{2}\). Since \(p = \frac{1}{2} < 1\), this series **diverges**.

6. **Conclusion**:  
   By the Limit Comparison Test, because \( \sum b_n \) diverges, \( \sum a_n \) also **diverges**.

### Final Answer:
\[
\sum_{n=2}^{\infty} a_n \text{ diverges by the Limit Comparison Test.}
\]

Would you like more detailed steps, or have any questions on specific parts?

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### Relative Questions
1. What is the Limit Comparison Test, and how does it work?
2. Why is it important to select a suitable comparison series \(b_n\)?
3. What are the conditions for a \(p\)-series to converge or diverge?
4. How does the behavior of the terms as \(n \to \infty\) affect the choice of \(b_n\)?
5. Can the Limit Comparison Test be used if the limit is zero or infinite?

### Tip
Always simplify \(a_n\) as much as possible to identify a comparable \(p\)-series for easier application of the Limit Comparison Test.

Use the Limit Comparison Test to determine whether the series \( \sum_{n=2}^{\infty} \frac{16n - 5}{4n \sqrt{n}} \) converges or diverges.

Explore a step-by-step solution to determine whether the series \( \sum_{n=2}^{\infty} \frac{16n - 5}{4n \sqrt{n}} \) converges or diverges using the Limit Comparison Test. Learn about key concepts such as the Limit Comparison Test, P-Series, and convergence and divergence criteria. Suitable for advanced high school and college students.

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