Let $a_n$ be a sequence that converges to $A$ with order of $n^alpha$, that is $a_n = A + mathcal{O}(n^alpha)$ and $b_n$ is another sequence that converges to B with ordder of $n^beta$; i.e. $b_n = B + mathcal{O}(n^beta)$. What is the order of convergence of $a_n cdot b_n$?

My intuition tells me that:

$$ a_nb_n = AB + mathcal{O}(n^{min(alpha,beta)})$$

but could not formulate the proof. I used the trick:

$$|a_nb_n – a_nB + a_nB – AB| leq |a_n||b_n-B| + |B| |a_n-A| leq K n^{alpha}n^{beta}+ Bn^{alpha}$$