We have these two hypothesis:
$$\forall\epsilon_1>0, \exists n_0 | n>n_o \implies |a_n|<\epsilon_1$$
$$|b_n|<M$$
where $M$ is the sequence bound.
Therefore, I've used hypothesis 2 to multiply both sides in the hypothesis 1 so we have:
$$\forall\epsilon_1>0, \exists n_0 | n>n_o \implies |a_n b_n|<\epsilon_1M$$
then if we choose $\epsilon = \epsilon_1 M$ we have:
$$\forall\epsilon>0, \exists n_0 | n>n_o \implies |a_n b_n|<\epsilon$$
Am I right?
Best Answer
Hint: Take any $\epsilon >0$ then $|a_n| < \frac{\epsilon}{M}$, whenever $n> n_0$, for some $n_0 \in \mathbb N$, where $M$ is such that $|b_n|\leq M$ then
$$|a_n b_n|= |a_n||b_n|<\frac{\epsilon}{M} M = \epsilon $$