# Proof of Sum Rule for Sequences

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

Proof
Suppose that $\varepsilon >0$. As $a_n\to a$ there exists an
$N_1\in \N$ such that for all $n>N_1$ we have $|a_n-a|\lt \varepsilon /2$. Similarly for $b_n$ given the same $\varepsilon$ there exists a $N_2$ such that $n>N_2 \Longrightarrow |b_n-b|\lt \varepsilon /2$.

For $n>\max \{ N_1,N_2\}$ we have
\begin{align*}
|(a_n+b_n)-(a+b)| &= |(a_n-a) + (b_n-b)| \\
&\leq |a_n-a | + |b_n-b| \\
&<\dfrac{\varepsilon }{2} +\dfrac{\varepsilon }{2} \\ &= \varepsilon. \end{align*} Hence, $a_n+b_n\to a+b$.