The problem asks about the convergence properties of the series:

$\sum_{n=1}^{\infty} \frac{\cos(n)}{n^2(n^{1/2})}$

This can be simplified as:

$\sum_{n=1}^{\infty} \frac{\cos(n)}{n^{5/2}}$

Analyzing the Series:

1. General Form: The series is of the form $\sum_{n=1}^{\infty} \frac{\cos(n)}{n^{5/2}}$. Notice that $|\cos(n)| \leq 1$, so the behavior of the series is dominated by $\frac{1}{n^{5/2}}$.

2. Comparison Test: The term $\frac{1}{n^{5/2}}$ resembles a p-series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ with $p = 5/2$. A p-series converges if $p > 1$. Since $5/2 > 1$, the series $\sum_{n=1}^{\infty} \frac{1}{n^{5/2}}$ converges.

3. Absolute Convergence: The absolute series would be $\sum_{n=1}^{\infty} \left|\frac{\cos(n)}{n^{5/2}}\right| \leq \sum_{n=1}^{\infty} \frac{1}{n^{5/2}}$, which is a convergent p-series as we have shown. Therefore, the original series converges absolutely.

Conclusion:

Since the series converges absolutely, the correct answer is:

a. Absolutely convergent

Would you like more details or explanations? Here are five related questions to expand on this concept:

1. What is the p-series test, and how does it determine convergence?
2. How does absolute convergence differ from conditional convergence?
3. Can the comparison test be used to prove divergence?
4. Why does the cosine function not affect the convergence of the series significantly in this case?
5. How does the behavior of $\cos(n)$ impact the terms of the series?

Tip: Absolute convergence implies both the original and absolute series converge, which is a stronger form of convergence.