real analysis – Order of convergence of a product of two convergent sequences

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:

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