Proving {$b_n$}$_{n=1}^\infty$ converges given {$a_n$}$_{n=1}^\infty$ and {$a_n b_n$}$_{n=1}^\infty$ Suppose {$a_n$}$_{n=1}^\infty$ and  {$b_n$}$_{n=1}^\infty$ are sequences such that {$a_n$}$_{n=1}^\infty$ coverges to A$\neq$0 and {$a_n b_n$}$_{n=1}^\infty$ converges.  Prove that {$b_n$}$_{n=1}^\infty$ converges.
What I have so far:
$b_n = {a_n b_n \over a_n}$ $\to$ $C \over A$, $A\neq0$
|$b_n - {C \over A}$| = |${a_n b_n \over a_n} - {C \over A}$| = |${Aa_nb_n - Ca_n \over Aa_n}$| $ \leq $ |${1 \over Aa_n}||Aa_nb_n - Ca_n$|=|${1 \over Aa_n}||a_n(Ab_n - C)$|
$\leq |{1 \over Aa_n}||a_n||(Ab_n - C)$|.  Note: since $a_n$ converges, there is M>0 such that |$a_n| \leq$M for all n $ \in\Bbb N$.
Thus, |${1 \over Aa_n}||a_n||(Ab_n - C)$| = |${1 \over M}||M||(Ab_n - C)$|.  And this is where I get lost.  Any thoughts? Or am I completely wrong to begin with?

Hint: Let $A=\lim\limits_{n\to\infty}a_n$ and $B=\frac{\lim\limits_{n\to\infty}a_nb_n}{A}$.
Since $|A|>0$, for $n$ large enough, $|a_n-A|\le\frac{|A|}{2}$. Show that then, $|a_n|\ge\frac{|A|}{2}$.
Then note that
$$
a_n(b_n-B)=(a_nb_n-AB)-B(a_n-A)
$$
