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?

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.