Proof of Product Rule for Sequences

$$\def \C {\mathbb{C}}\def \R {\mathbb{R}}\def \N {\mathbb{N}}\def \Z {\mathbb{Z}}\def \Q {\mathbb{Q}}\def\defn#1{{\bf{#1}}}$$
Product Rule
Suppose that $(a_n)$ and $(b_n)$ are two convergent sequences with $a_n\to a$ and $b_n\to b$. Then
$a_n b_n \to ab$.

Proof
Given $\varepsilon >0$, there exists $N_1$ and $N_2$ such that

\begin{align*}
|a_n-a|&<\sqrt{\varepsilon /3} {\text{ for all }} n>N_1\\
{\text{and }} |b_n-b|&<\sqrt{\varepsilon /3} {\text{ for all }} n>N_2.
\end{align*}
There exists $L,M\in \R$ such that $|a|\lt L$ and $|b|\lt M$. This implies that there exist $N_3$ and $N_4$ such that
\begin{align*}
|a_n-a|&<\dfrac{\varepsilon}{3M} {\text{ for all }} n>N_3\\
{\text{ and }} |b_n-b|&<\dfrac{\varepsilon}{3L} {\text{ for all }} n>N_4.
\end{align*}
Let $N=\max \{N_1,N_2,N_3,N_4\}$. Then, for all $n>N$,
\begin{align*}
\left| a_nb_n-ab\right|&=\left|(a_n-a)(b_n-b) + a (b_n-b)+b(a_n-a) \right|\\
&\leq |a_n-a||b_n-b| + |a| |b_n-b| + |b||a_n-a| \\
&\lt |a_n-a||b_n-b| + L| |b_n-b| + M|a_n-a| \\
&\lt \sqrt{\dfrac{\varepsilon}{3}} \sqrt{\dfrac{\varepsilon}{3}} + \dfrac{\varepsilon}{3} + \dfrac{\varepsilon}{3} \\
&=\varepsilon .
\end{align*}



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