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?
 A: Mh what is about something like:
$a_n b_n$ converges now let's call the limit $C$, as $A\neq 0$ we can write 
$C=A\cdot B$ with $B=\frac{C}{A}$ 
$$
\begin{align*}
 |a_n b_n - A \cdot B| &= |a_n b_n - b_n A +b_n A - A\cdot B|\\
&=|b_n(a_n-A) + A(b_n-B)| \\
\end{align*} $$
Because $a_n$ converges the first part converges to zero, as we know the lhs converges the rhs has to converges to, so 
$$|A(b_n-B)|=|A| |b_n-B| $$ 
must converge to zero, as we know $|A|\neq 0$ we know 
$$ |b_n-B|$$ converges to $0$.  And so $b_n$ converges to $B$
As robjohn pointed out we get the triangle inequality 
$$|A(b_n-B)|\leq |a_n b_n - AB| + |b_n(a_n-A)| $$ As we know $a_nb_n$ converges with $a_n$ not converging to $0$. If $b_n$ is not bounded $a_n b_n$ can't converge, as we know $a_nb_n$ is convergent, we get $b_n$ is bounded.
A: 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
$$
a_n(b_n-B)=(a_nb_n-AB)-B(a_n-A)
$$
A: $$\frac{\lim_{n\to\infty}a_n b_n}{\lim_{n\to\infty}a_n} = \lim_{n\to\infty}\frac{a_n b_n}{a_n} =\lim_{n\to\infty}b_n.$$
This is legitimate because you start with limits that you know exist.  The rest is standard properties of limits of sequences found in any calc textbook.
