If $(a_n)$ is a null sequence and $(b_n)$ is bounded, then $(a_nb_n)$ is a null sequence

Solution 1:

I like to think of proofs like this as challenge/response. If you claim $a_n$ is null, I can challenge you with any $\epsilon \gt 0$ and you have to be able to find an $N$ such that ...

Now you are claiming that if I challenge you with some $\epsilon_2$, you can find an $N_2$ such that $a_nb_n \lt \epsilon_2$ as long as $n \gt N_2$. Somebody told you that $a_n$ was null. Can you find an $\epsilon_3$ to challenge him with and use the $N_3$ that comes back?

Solution 2:

HINT: For all $n\in\Bbb N$, $|a_nb_n|\le A|a_n|$. Use the fact that $\langle a_n:n\in\Bbb N\rangle$ is null to show that $\langle Aa_n:n\in\Bbb N\rangle$ is null, and then squeeze.