The inequality involving $\frac{1}{\sqrt{x} + e^{3x}} \leq \frac{1}{\sqrt{x}}$ is not helpful for proving convergence because the integral $\int_1^\infty \frac{1}{\sqrt{x}} \, dx$ diverges. This happens because $\int_1^\infty \frac{1}{\sqrt{x}} \, dx$ is a $p$-integral with $p = \frac{1}{2}$, and $p < 1$, which guarantees divergence.

Using a divergent comparison (like $\frac{1}{\sqrt{x}}$) does not allow us to conclude anything about the convergence or divergence of the given integral $\int_1^\infty \frac{1}{\sqrt{x} + e^{3x}} \, dx$.

Instead, the proper comparison is made with $\frac{1}{e^{3x}}$, as the larger function $\int_1^\infty \frac{1}{e^{3x}} \, dx$ converges, allowing the use of the Comparison Test to conclude that the original integral also converges.

Would you like further clarification on this concept or detailed steps on how the $p$-integral divergence works?

Here are 5 related questions to deepen understanding:

1. Why does $\int_1^\infty \frac{1}{\sqrt{x}} \, dx$ diverge, and how does the $p$-integral test explain it?
2. What are the criteria for the Comparison Test to determine the convergence of an integral?
3. How does exponential decay $e^{3x}$ ensure convergence of the integral $\int_1^\infty \frac{1}{e^{3x}} \, dx$?
4. Could we apply the Limit Comparison Test instead of the direct Comparison Test? Why or why not?
5. How would the behavior of $\sqrt{x} + e^{3x}$ change if $x \to 0^+$ rather than $x \to \infty$?

Tip: When using the Comparison Test, always compare with a simpler function that you know converges or diverges, and ensure that the comparison is valid for all $x$ in the range of the integral.