A step in proving that a real Cauchy sequence is convergent. I'm trying to prove that a real Cauchy sequence is convergent, but I need some help for a step.
We have the following statements:


*

*$\{ s_i\}$ is a real Cauchy sequence, i.e. $\forall\epsilon>0,\exists N\geq 1,\forall n\geq N,\forall k\geq 1, \mid s_i-s_{i+k}\mid<\epsilon$

*Each $s_i$ is an equivalence class of rational Cauchy sequence,

*For each $i$, we define $\epsilon_i=\frac{1}{2i}$,

*For each $i$, we choose $N_i$ using the definition of a Cauchy sequence
$$\exists N_i\geq1,\forall n\geq N_i,\forall k\geq 1, \mid s_{i,n}-s_{i,n+k}\mid\leq\frac{1}{2i}:=\epsilon_i$$

*We define $v_i:=s_{i,N_i}$

*We already shown that 


*

*$\mid v_i -s_i\mid<\frac{1}{i}$

*$\{v_i\}$ is a rational Cauchy sequence


*We define $s$ as an equivalence class $s:=\overline{\{v_n\}}$ 
Now I need to show that $\mid v_i -s\mid<3\epsilon$
--------EDIT--------
We can show that $\mid v_i-v_{i+k}\mid<2\epsilon$ with
$\mid v_i-v_{i+k}\mid=\mid v_i-s_i+s_i-s_{i+k}+s_{i+k}-v_{i+k}\mid$
$\leq\mid v_i-s_i\mid+\mid s_i-s_{i+k}\mid+\mid s_{i+k}-v_{i+k}\mid<\frac{1}{i}+\epsilon+\frac{1}{i+k}<2\epsilon$

It's important to keep the distinction between equivalence class and reprentative clear to show this.
The real number $s$ is an equivalence class of rational Cauchy sequences that are equivalent to $\{v_n\}$.  Let $\{q_n\}$ be any representative. Then $\{q_n\} \sim \{v_n\}$ and given $\epsilon > 0$ we have $|q_n - v_n| < \epsilon$ for $n \geq N'$.
The rational number $v_i$, considered as a real number,  is the equivalence class of the sequence $(v_i,v_i,v_i, \ldots).$  Hence, $|v_i - s|$ is a real number with a representative Cauchy sequence $\{|v_i - q_n|\}.$
Then for $n \geq N'$
$$|v_i - q_n| \leqslant |v_i - v_n| + |v_n - q_n| < \epsilon + |v_i - v_n|.$$
Also $\{v_i\}$ is itself a rational Cauchy sequence with $|v_i - v_{i+k}| < 2\epsilon$, as shown in the EDIT, for $i$ sufficiently large. Then $|v_i -v_n| < 2 \epsilon$ if $n > \max(N',i)$. 
Therefore, every representative of $|v_i - s|$ satsifies $|v_i - q_n| < 3\epsilon$ for sufficiently large $i$. 
