I have two integrals:
$ A=\int\limits_1^\infty \dfrac{1}{x}dx\,, B=\int\limits_1^\infty \dfrac{1}{x^2}dx $
Calculus says that A is an improper integral as it diverges, but the B converges and is $1$, because $\dfrac{1}{x^2}$ is faster near $y=0$ than $\dfrac{1}{x}$.
I don't understand the reason behind this. So I looked for another way to put down my problem. Multiple sources define that:
$ \frac{1}{\infty} = \frac{1}{\infty^2} $
What is the reason that the integral of $\dfrac{1}{x}$ is divergent and $\dfrac{1}{x^2}$ is convergent? In the end they both reach $\dfrac{1}{\infty}$ (or $\dfrac{1}{\infty^2}$ which is $\dfrac{1}{\infty}$).