Show that in $\mathbb{R}^n$, sum of two Cauchy sequences is Cauchy. To show it I started off from the following:
Let $(x_n)$ be a Cauchy sequence in $\mathbb{R}^n$ so 
$$
\forall \,\,\,\,\epsilon_1 >0\,\,\,\,\exists \,\,\,N_1 \in \mathbb{N}\,\,\,\ \text{such that} \,\,\,\, n,m \geq N_1 \,\,\,\, \rightarrow d(x_n,x_m) <\epsilon_1
$$
Also Let $(y_n)$ be a Cauchy sequence in $\mathbb{R}^n$ so 
$$
\forall \,\,\,\,\epsilon_2>0 \,\,\,\,\exists \,\,\,N_2 \in \mathbb{N}\,\,\,\ \text{such that} \,\,\,\, p,q \geq N_2 \,\,\,\, \rightarrow d(y_p,y_q) <\epsilon_2
$$
Let $N=\max(N_1,N_2),r=\max(n,p),s=\max(m,q)$
Then
$$
\forall \,\,\,\,\epsilon_1>0 \,\,\,\text{and}\,\,\,\,\epsilon_2>0 \,\,\,\,\exists \,\,\,N \in \mathbb{N}\,\,\,\ \text{such that} \,\,\,\, r,s \geq N \,\,\,\, 
$$
$$
\rightarrow d(x_r,x_s) < \epsilon_1 \,\,\,\text{and} \,\,\,\,d(y_r,y_s)< \epsilon_2
$$
I do not know how to show $(x_n+y_n)$ be Cauchy?
If we sum them we have 
$$
\rightarrow d(x_r,x_s) +d(y_r,y_s) < \epsilon_1 + \epsilon_2
$$
But I need
$$
d(x_r+y_r,x_s+y_s) < \epsilon_1 + \epsilon_2=\epsilon
$$

Given $\epsilon$, there exists $N_1\in\Bbb N$ such that when $r,s\ge N_1$ we have that $|x_r-x_s|<\dfrac{\epsilon}{2}.$ Similarly, there exists $N_2\in\Bbb N$ such that $r,s\ge N_2$ implies $|y_r-y_s|<\dfrac{\epsilon}{2}.$
Now let $N=\max\{N_1,N_2\}$. So for $r,s\ge N$, we have
$$|(x_r+y_r)-(x_s+y_s)|=|(x_r-x_s)+(y_r-y_s)|<\dfrac{\epsilon}{2}+\dfrac{\epsilon}{2}=\epsilon$$
