How can you prove that $(a_nb_n)$ is a null sequence given that $(a_n)$ is a null sequence that converges to zero and $(b_n)$ is bounded above by $A$?
The conditions of $(a_n)$ are: For every $\varepsilon > 0$ there exists an $N$ in Natural numbers such that $|a_n| < \varepsilon$ for all $n \ge N$.
The conditions of $(b_n)$ are: $b_n$ is a function natural numbers -> reals such that there exists $A$ with $|b_n| \le A$ for all $n$ in natural numbers.
Best Answer
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?