# Convergence of product of sequence and convergence of sequence

Suppose if we have bounded, non-negative sequences $$\{a_n\},\{b_n\}$$ such that $$a_n\to A >0$$. Then is it true that, if $$\{a_nb_n\}$$ converges then $$\{b_n\}$$ must also converge? My intuition is yes, since $$a_n$$ does not converge to $$0$$. I try to prove it here:

Suppose towards a contradiction that $$\{b_n\}$$ does not converge, then one can obtain two subsequences that converge to different limits, say $$b_{n_j}\to L_1$$ and $$b_{n_k}\to L_2$$, where $$L_1\neq L_2$$. But then, as $$a_n$$ converges, we must have $$a_{n_j}b_{n_j}\to A(L_1)$$ and $$a_{n_k}b_{n_k}\to A(L_2)$$. Since $$A>0$$, we must have $$A(L_1)\neq A(L_2)$$ and thus the product of the sequence does not converge.

Please let me know if this proof makes sense as I couldn't find a reference. Thanks!

Your proof is correct but I would argue as follows: $$b_n=(a_nb_n)\frac 1 {a_n} \to \frac L A$$ where $$L=\lim a_nb_n$$.