Proving null sequences

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.

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What is a "null sequence?" I assumed it meant converges to $0$, but then you write, "given that $a_n$ is a null sequence that converges to zero," which would be redundant if that was the meaning of "null sequence." – Thomas Andrews Oct 23 '12 at 19:36
You are almost done - from things you wrote you get that $|a_nb_n|\le A\varepsilon$ whenever $n\ge N$. – Martin Sleziak Oct 23 '12 at 19:39
@ThomasAndrews Yes, that is what Martin meant. – AD. Oct 23 '12 at 19:41
Ah, and he edited it, so I'll delete my comment :) – Thomas Andrews Oct 23 '12 at 19:42

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.
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?
How do I challenge your claims when I am not told what the sequences actually are? Therefore I cannot find such N when you challenge me with $\epsilon>0$? – Matt Oct 23 '12 at 20:09
You are given that $a_n$ is null, so challenge the person with $\frac \epsilon A$. What can you do with the N that comes back? – Ross Millikan Oct 23 '12 at 20:33