Multiply a Known Convergent Sequence by a Bounded Sequence I have been tasked with proving the following:

$a_n$ is bounded but not necessarily convergent
assume that $\lim{b_n} \to 0$
show $\lim{a_n b_n} \to 0$.

I started my proof listing what I know:

*

*If $a_n$ is bounded then there is some number $M \gt 0$ such that $|a_n| \le M$ for all $n \in \mathbb{N}$.

*$|b_n - 0| \lt \epsilon$.

My proof proceeded directly from this:
$|a_n b_n| = |a_n| |b_n| \lt \epsilon$
$|b_n| \lt \frac{\epsilon}{|a_n|}$
And since $|b_n - 0| = |b_n|$ can be made arbitrarily small by (2) above we have shown that for all $n \ge N$ ($N \in \mathbb{N}$),
$|a_n b_n| = |a_n| |b_n| \lt |a_n| \frac{\epsilon}{|a_n|} = \epsilon$.
Our limit therefore converges to 0 as desired.
I was fairly satisfied with this proof, I had considered $M$ being able to be replaced by $|a_n|$ (as I didn't see a reason I couldnt do this), so this took advantage of (1) as well.
The author went about it in a different but similar way starting from the same "base":
$|a_n b_n - 0| = |a_n| |b_n| \le M|b_n|$ from (1)
Because $(b_n) \to 0$ we can pick an $N$ such that:
$|b_N| \lt \frac{\epsilon}{K}$
Finally, we can conclude that for this choice of $N$,
$|a_n b_n - 0| \lt K |b_n| \lt K \frac{\epsilon}{K} = \epsilon$
The author seemed to have just "dropped" the $|a_n|$ from the last part where he does $K |b_n|$?
Other than that weirdness (and the ham fisted nature of my newbie proof), is there any fundamental differences in our approach? Am I not able to let $M = |a_n|$?
 A: Your proof is fine, but it's hard to read.
Here's a revised version:
Let $M$ be a positive number such that $|a_{n}|\leq M$ for all $n$.
Then,
$$
\left|a_{n}b_{n}\right|
=\left|a_{n}\right|\left|b_{n}\right|
=M\left|b_{n}\right|.
$$
For each $\epsilon>0$, pick $N$ large enough such that $|b_{n}|<\epsilon/M$ whenever $n\geq N$.
It follows that $|a_{n}b_{n}|<M(\epsilon/M)=\epsilon$ whenever $n\geq N$.
Therefore, $a_{n}b_{n}\rightarrow0$.
A: Your mistake is when you put $\dfrac{\epsilon}{\vert{a_n} \vert}$ this expression depends on $n$ so your inequality is only valid for that $n$.
When you use the fact that $\vert a_n \vert \leq M, \forall n \in \mathbb{N}$ then that number $M$ can be used at your convenience to find values ​​that meet the definition of limit.
What you want to have is that $\vert a_n b_n \vert < \epsilon$ for very large values ​​of $n$. So if you give some $ \epsilon >0$ the value $\dfrac{\epsilon}{M} >0$ and applying the definition of limit you should that : $\exists$ $n_0 \in \mathbb{N} / n>n_0 : \vert b_n \vert < \dfrac{\epsilon}{M}$.
Now for $n>n_o : \vert a_n b_n \vert = \vert a_n \vert \vert b_n \vert < M. \dfrac{\epsilon}{M} = \epsilon $ and you have what you were looking for.
