In order to solve this problem, I am going to assume that $b_{n}\rightarrow+\infty$.
Since $a_{n}$ converges, it is bounded. That is to say $|a_{n}|\leq M$.
On the other hand, for every $M/\varepsilon > 0$, there exists $N \geq 0$ such that
\begin{align*}
n\geq N \Longrightarrow b_{n} \geq \frac{M}{\varepsilon} \Longrightarrow \frac{1}{|b_{n}|} \leq \frac{\varepsilon}{M}
\end{align*}
Consequently, for every $\varepsilon > 0$, there exists $N \geq 0$ such that
\begin{align*}
n\geq N \Longrightarrow \left|\frac{a_{n}}{b_{n}} - 0\right| \leq M\times\frac{\varepsilon}{M} = \varepsilon
\end{align*}
In order to solve this problem, I am going to assume that $b_{n}\rightarrow+\infty$.
Since $a_{n}$ converges, it is bounded. That is to say $|a_{n}|\leq M$.
On the other hand, for every $M/\varepsilon > 0$, there exists $N \geq 0$ such that \begin{align*} n\geq N \Longrightarrow b_{n} \geq \frac{M}{\varepsilon} \Longrightarrow \frac{1}{|b_{n}|} \leq \frac{\varepsilon}{M} \end{align*}
Consequently, for every $\varepsilon > 0$, there exists $N \geq 0$ such that \begin{align*} n\geq N \Longrightarrow \left|\frac{a_{n}}{b_{n}} - 0\right| \leq M\times\frac{\varepsilon}{M} = \varepsilon \end{align*}
Consequently $a_{n}/b_{n}\to 0$, and we are done.