I'm trying to prove, given only that one sequence $B$ is bounded and the other sequence $A$ is convergent to zero, that their product also converges.
My work probably wouldn't be considered a formal proof but I'm curious if my strategy/thought process is on the right track.
My work:
Since $A$ converges to zero, there exists a point in this sequence where all terms are between $-1$ and $1$. In this case, whatever behavior $B$ has will be dictated by the terms of $A$. If $B=0$, then the conclusion is trivial. Thus, $AB$ converges to zero.
Here's a formal one: Let $M>0$ be such that $|y_{n}|\leq M$. Let $\{x_{n}\}$ be such that $x_{n}\rightarrow 0$. Given $\epsilon>0$, for the positive number $\epsilon/M$, there exists some $N$ such that $|x_{n}|<\epsilon/M$ for all $n\geq N$, so $|x_{n}y_{n}|\leq M|x_{n}|<M\cdot\epsilon/M=\epsilon$ for all such $n$, this shows that $x_{n}y_{n}\rightarrow 0$.