I know the comparison test for improper integrals says:
"If $f(x), g(x)$ are continuous such that $0\leq g(x)\leq f(x)$ for $x\geq a,$ and if $\int_{a}^\infty{f(x)dx}$ converges, then $\int_{a}^\infty{g(x)dx}$ converges.
If $\int_{a}^\infty{g(x)dx}$ diverges, then $\int_{a}^\infty{f(x)dx}$ diverges.
However, when we see that $f(x)$ diverges, we can't say anything about the convergence of $g(x).$ What are we supposed to do in this situation?
Edit: I have been told sometimes that you can set $g(x)$ as an upper bound and search for a new function $h(x) \leq g(x).$ If $h(x)$ diverges, then $g(x)$ must diverge. However, I don't think this makes sense because if $h(x)$ converges, you are stuck again.
The comparison test for integrals can only tell you if certain integrals converge, given certain bounds. If f(x) $\le$ g(x) for on some interval I, then $$\displaystyle\int_I f(x) dx \le \displaystyle\int_I g(x) dx$$
So if f diverges, then g must diverge and the contrapositive is true: that is if g converges, then f converges. Those are the only conclusions to be drawn. the hypotheses are not satisfied, then you have to use a different convergence test OR different functions.