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!


1 Answer 1


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


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