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.

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    $\begingroup$ Your proof is correct. I think that notation would be lighter if you call your limits $\lim_{n \to \infty} a_n = A$ and $\lim_{n \to \infty} b_n = B$, however this is just a matter of taste. $\endgroup$ – Crostul Mar 31 '15 at 11:22
  • $\begingroup$ I assume the proof doesn't require any differences to make it apply to pointwise or uniform convergent sequences. $\endgroup$ – mavavilj Sep 30 '16 at 16:40

Your proof is correct. $ { } $

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