Lebl-Basic-Analysis-I · Problem 5.5.12

The key insight is that boundedness of $g(x) = x^2 f(x)$ gives you a powerful control on how fast $f(x)$ decays—it must shrink at least as quickly as $1/x^2$. Since $\int_1^{\infty} \frac{1}{x^2} dx$ converges, you can use the comparison test to pin down convergence of $\int_1^{\infty} f$. The bounded function $g$ is your bridge between the abstract integral and a concrete, well-behaved comparison.

The key insight is that boundedness of $g(x) = x^2 f(x)$ gives you a powerful control on how fast $f(x)$ decays—it must shrink at least as quickly as $1/x^2$. Since $\int_1^{\infty} \frac{1}{x^2} dx$ converges, you can use the comparison test to pin down convergence of $\int_1^{\infty} f$. The bounded function $g$ is your bridge between the abstract integral and a concrete, well-behaved comparison.