$b_n \to +\infty$, $b_n > 0$, $n \to \infty$. Prove $a_n$ is convergent to $0$ if $a_n b_n$ is bounded. $b_n \to +\infty$, $b_n > 0$, $n \to \infty$. Prove $a_n$ is convergent to $0$ if $a_n b_n$ is bounded.
There are a lot of questions like "$a_n$ is convergent, $b_n$ is bounded. Prove $a_n b_n$ is convergent/bounded/evaluate lim etc".
So my questions is, I guess, pretty basic, but a little bit different (at least I hope so)
Just to clarify, the question is why if $a_n b_n$ is bounded, while $b_n \to +\infty$, implies ($\implies$) that $a_n \to 0$
Intuitive thinking is the following: if $b_n \to +\infty$, then $a_n$ must be decreasing (at least as fast as $b_n$ growth).  So, for example, $a_n = \frac{k}{b_n}$, $k$ -- some fixed number. Hence, $a_n \to 0$, $\lim_{n \to \infty} \frac{k \times b_n}{ b_n} = k$.
Unfortunately, it can't be counted as proof.  Also I'm not so good at $\epsilon$-notation.
But, my guess, we must somehow express $a_n$ through $b_n$.
So I've got the idea "what to do" without knowledge "how to do".
 A: Hint:
$$
   |a_n | = |a_n b_n| \left|\frac{1}{b_n}\right|
$$
Since $(a_nb_n)$ is bounded, there exists $M\geq 0$ such that $|a_nb_n| \leq M$ for all $n$. Since $\frac{1}{b_n} \to 0$, for all $\epsilon > 0$, there exists $N$ such that $\left|\frac{1}{b_n}\right| < \epsilon$ whenever $n \geq N$. Since $ M$ is fixed and $\epsilon$ is arbitrary, you can make $M \epsilon$ as small as necessary.
Does that help?
A: $b_n \to +\infty$ means that for every $A>0$ there is $N \in \mathbb{N}$ so that $b_n>A$ whenever $n \geq N$.
$\{a_nb_n\}$ bounded means that there is $L>0$ so that $|a_nb_n| \leq L$ for every $n \in \mathbb{N}$.

Let $\varepsilon>0$ be given. Since $\{a_nb_n\}$ is bounded, there is $L>0$ so that $|a_nb_n| \leq L$ for every $n \in \mathbb{N}$. Set $A=\max\{L/\varepsilon,L\}$. Since $A>0$, there is $N \in \mathbb{N}$ so that $b_n>A$ whenever $n \geq N$. We combine to conclude
\begin{align*}
|a_n| \leq |a_n|\frac{b_n}{A}=\frac{1}{A}|a_nb_n| \leq \frac{L}{A} \leq \varepsilon \text{ whenever } n \geq N.
\end{align*}
A: Your intuition is good.
$a_nb_n$ is bounded means that there are upper bound $U$ so that $L< a_nb_n < U$ for all $n$.  That would mean $\frac L{b_n} < a_n < \frac U{b_n}$ for all $n$.  But as $n\to \infty$ we have $b_n \to \infty$ we have $\frac U{b_n}\to 0$ and $\frac L{b_n} \to 0$ so by the squeeze thereom $a_n \to 0$.
To get practice with $N, \epsilon$ proofs:
For every $\epsilon > 0$ we need to find an $N$ so that $n > N$ would imply $|a_n| < \epsilon$.  How can we find such an $N$?
We have by the argument above $\frac L{b_n} < a_n < \frac U{b_n}$ (Note: we have no idea whether $L$ or $U$ are pos or negative) so $|a_n| < \max(\frac {|L|}{b_n}, \frac {|U|}{b_n})$.  So if we can assure that $|a_n| < \max(\frac {|L|}{b_n}, \frac {|U|}{b_n}) <\epsilon$ we would be done.  Can we do that.
Well for  $\max(\frac {|L|}{b_n}, \frac {|U|}{b_n}) <\epsilon$ that would occur if $b_n > \frac {\max(\frac{|L|,|U|}\epsilon$.
Well, what do we know about $b_n$.  We know that $\lim_{n\to\infty} b_n = \infty$.  And what does that literally mean?  It means for any $K \in \mathbb R$ there is an $N$ so that if $n> N$ then $b_n > K$.
So if we let $\frac {\max(|L|,|U|)}\epsilon$ be the $K$ then there is an $N$ so that $n > N$ would imply $b_n >\frac {\max(|L|,|U|)}\epsilon$.
And that's it, we are done, as that exact same $N$ is enough to show that
$n > N\implies $
$b_n > \frac {\max(|L|,|U|)}\epsilon \implies$
$\epsilon > \frac {\max (|L|, |U|)}{b_n} > |a_n|$
So $\lim_{n\to \infty} a_n = 0$.
