Does my proof appear correct? Also, do you like the notation?
$\textbf{Theorem.}$
If
$(a_n)_{n \in \mathbb{N}}$ and
$(b_n)_{n \in \mathbb{N}}$
are convergent real sequences, then
$$
\lim_{n \to \infty} \left( a_n + b_n \right) =
\left( \lim_{n \to \infty} a_n \right) +
\left( \lim_{n \to \infty} b_n \right) .
$$
$\textit{Proof.}$
Let $\varepsilon > 0$.
It remains to prove that
there is $N \in \mathbb{N}$ such that,
for every natural number $m > N$,
we have
$$
\left|
a_m + b_m –
\left( \lim_{n \to \infty} a_n \right) –
\left( \lim_{n \to \infty} b_n \right)
\right| < \varepsilon .
$$
By assumption,
there is $N_a \in \mathbb{N}$ such that,
for every natural number $m > N_a$,
we have
$$
\left|
a_m –
\left( \lim_{n \to \infty} a_n \right)
\right| < \varepsilon / 2 .
$$
Analogously,
there is $N_b \in \mathbb{N}$ such that,
for every natural number $m > N_b$,
we have
$$
\left|
b_m –
\left( \lim_{n \to \infty} b_n \right)
\right| < \varepsilon / 2 .
$$
We choose $N := \max(N_a, N_b)$.
Let $m \in \mathbb{N}$ such that $m > N$.
Obviously, for this $m$, each of the above two inequalities holds.
Thus, we may add the inequalities. Doing so, we obtain
\begin{equation*}
\begin{split}
\varepsilon = \varepsilon / 2 + \varepsilon / 2
& >
\left|
a_m –
\left( \lim_{n \to \infty} a_n \right)
\right|
+
\left|
b_m –
\left( \lim_{n \to \infty} b_n \right)
\right| \\
& \ge
\left|
a_m –
\left( \lim_{n \to \infty} a_n \right) +
b_m –
\left( \lim_{n \to \infty} b_n \right)
\right| &\quad& \text{by subadditivity of abs. val.} \\
& =
\left|
a_m + b_m –
\left( \lim_{n \to \infty} a_n \right) –
\left( \lim_{n \to \infty} b_n \right)
\right| .
\end{split}
\end{equation*}
Hence, by transitivity,
\begin{equation*}
\varepsilon >
\left|
a_m + b_m –
\left( \lim_{n \to \infty} a_n \right) –
\left( \lim_{n \to \infty} b_n \right)
\right|,
\end{equation*}
QED.
Best Answer
Your proof is correct. $ { } $