Properties of Big $\mathcal{O}$

Question

I have seen in a paper that, if $A=\mathcal{O}(p^2)$ and $B=\mathcal{O}(p)$ then,

how can we say that, $A^{-1/2}B$ is diverging?

The way I thought is,

if $A = \mathcal{O}(p^2)$, then $A^{-1/2}$ = $\mathcal{O}(p^{-1})$, then $A^{-1/2}B$ = $\mathcal{O}(p^{-1})\mathcal{O}(p) = \mathcal{O}(1)$. If so, we can't say it is diverging?

Any help is greatly appreciated.

Answer 1

We say that $f = O(g)$ (as $x \to \infty$) if there exists a constant $M > 0$ and a point $x_0 \in \mathbb{R}$ such that $$|f(x)| \le M | g(x) |$$ for all $x > x_0$. This just gives us upper bounds. Take for instance $A(p) = 1$ (a constant function) and $B(p) = 1/p$, then trivially we have $A = O(p^2)$ and $B = O(p)$. However $A^{-1/2} B = 1/p$, and converges to 0 as $p \to \infty$.

Answer 2

Thanks for the reference. Well, they state that if $A = \mathcal{O}(p^2)$ then $\mathcal{O}(p\cdot A^{-1/2})$ can diverge. Take a look at Grandi's serie which is $1-1+1-1\ldots$ which is in $\mathcal{O}(1)$ but does diverge.