is there a short, simple proof of the equivalence. I need just a direction from convergence to Cauchy, since the other direction I've already done and the proof , that I found here:
https://proofwiki.org/wiki/Cauchy_Sequence_Converges_on_Real_Number_Line
is rather complex. Thank you
On
Hint:-
$$\displaystyle\lim_{n\to\infty}a_n=l \implies \forall\epsilon>0, \exists n_0\mid \forall n\geq n_0, \left\lvert a_n-l\right\rvert<\epsilon$$
$\left\lvert a_{n+p}-a_n\right\rvert=\left\lvert (a_{n+p}-l)+(l-a_n)\right\rvert \leq \left\lvert a_{n+p}-l\right\rvert+\left\lvert l-a_n\right\rvert$
After establishing that $<a_n>$ has a convergent subsequence $<a_{n_k}>$, let $\displaystyle L = \lim_{k \to \infty} a_{n_k}$ and let $\varepsilon > 0$. There exists an $r \in \Bbb N$ such that $|a_{n_k} - L| < \frac{\varepsilon}{2}$ for all $k \ge r$. Since $<a_n>$ is Cauchy, there exists an $s \in \Bbb N$ such that $|a_n - a_m| < \frac{\varepsilon}{2}$ for all $n, m \ge s$. Set $j = \max\{r, s\}$. If $n \ge j$, then $|a_n - a_{n_j}| < \frac{\varepsilon}{2}$ (since $n \ge s$ and $n_j \ge j \ge s$) and $|a_{n_j} - L| < \frac{\varepsilon}{2}$ (since $j \ge r$). Thus, by the triangle inequality, $$|a_n - L| \le |a_n - a_{n_j}| + |a_{n_j} - L| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.$$ Since $\varepsilon$ is arbitrary, $<a_n>$ converges to $L$.