To solve this problem using the integral test, we analyze the series:

$\sum_{n=1}^{\infty} \frac{3}{\sqrt{n}}.$

Steps:

1. Define $f(x) = \frac{3}{\sqrt{x}}$, a continuous, positive, and decreasing function for $x \geq 1$.
2. Apply the integral test: Compute $\int_{1}^{\infty} \frac{3}{\sqrt{x}} \, dx$.

Integral Calculation:

$\int_{1}^{\infty} \frac{3}{\sqrt{x}} \, dx = \int_{1}^{\infty} 3x^{-1/2} \, dx = \left[ 6x^{1/2} \right]_{1}^{\infty}.$ At the upper limit ($x \to \infty$): $6\sqrt{x} \to \infty.$

Conclusion:

The improper integral diverges, so by the integral test, the series diverges.

Would you like further clarification or a step-by-step breakdown?
Here are 5 related questions for deeper understanding:

1. What are the conditions for applying the integral test?
2. How can you determine whether a function decreases on a given interval?
3. What are other convergence tests for series?
4. How does $\frac{1}{\sqrt{n}}$ compare to the $p$-series?
5. Why does divergence of the integral imply divergence of the series?

Tip: Before applying the integral test, ensure the function is continuous, positive, and decreasing.