My textbook, about the comparison test, said: if $f(x)\leq g(x)$, and $\int_a^\infty g(x)dx$ converges then $\int_a^\infty f(x)dx$ converges. But what if the integrals at hand are $\int_\infty^a$ bounded. Does this mean that $f(x)\geq g(x)$ for the test to hold? How do you show this?
I was specifically thinking about this due to trying to show the convergence of $\int_\infty^0 \frac{\sqrt{x}}{(x+1)^2}$ using $\int_\infty^0 \frac{\sqrt{x}}{x^2}$. If I flip the bounds, the functions will be in the negative form, and the second integral is greater in absolute value, as in more negative. If I am completely wrong about this, how can I show it?
Your situation involves taking the inequality deriving from the comparison test that you were talking about:
$$\int_a^\infty f(x) dx \leq \int_a^\infty g(x) dx$$
For $f(x) \leq g(x)$.
“Divide” both sides by -1, and consequently, switch the inequality symbol from $\leq$ to $\geq$.
$$\frac{1}{-1}\int_a^\infty f(x)dx \geq \frac{1}{-1}\int_a^\infty g(x)dx $$
With $\frac{1}{-1} = -1$ and using this to flip the bounds on both sides,
$$\int_\infty^a f(x)dx \geq \int_\infty^a g(x)dx $$
So this explains your concern hopefully.