$b_n \to +\infty$, $b_n > 0$, $n \to \infty$. 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".

$b_n \to +\infty$ 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*}
