$$\def \C {\mathbb{C}}\def \R {\mathbb{R}}\def \N {\mathbb{N}}\def \Z {\mathbb{Z}}\def \Q {\mathbb{Q}}\def\defn#1{{\bf{#1}}}$$

**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, $a_n+b_n\to a+b$.

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