Originally Posted by
ThePerfectHacker The sequence, $\displaystyle \{ a_n \}$ is a Cauchy sequence. Thus, for $\displaystyle n,m\geq N_1$ we have,
$\displaystyle |a_n-a_m| < \epsilon$ for any $\displaystyle \epsilon>0$
The sequence, $\displaystyle \{ b_n\} $ is a Cauchy sequence. Thus, for $\displaystyle n,m \geq N_2$ we have, $\displaystyle |b_n-b_m|< \epsilon $ for any $\displaystyle \epsilon>0$
Consider the sequence, $\displaystyle \{a_n+b_n\}$. For any $\displaystyle \epsilon /2>0$ we can find $\displaystyle N_1,N_2$ satisfy those two conditions on top. Let $\displaystyle N=\max\{N_1,N_2\}$ then,
$\displaystyle |a_n-a_m|<\epsilon/2$ and,
$\displaystyle |b_n-b_m|<\epsilon/2$.
Addition yields,
$\displaystyle |a_n-a_m|+|b_n-b_m|<\epsilon$
Triangular inequality (Thanks to CaptainBlank for teaching this move to me).
We have,
$\displaystyle |a_n+b_n-(a_m+b_m)|<\epsilon$ for $\displaystyle n,m\geq N$.
Q.E.D.