Prove every convergent sequence is Cauchy.

Question

I'm not sure how to go about this. I know if a sequence $\{a_n\}$ converges to $L$, then given $\epsilon > 0$:

(1) $\exists N$ such that for $n \geq N$, $|a_n - L| < \epsilon$.

Somehow, I must show:

(2) for large $n,m$, $|a_m - a_n| < \epsilon$.

I think by (1), we can choose $m > n \geq N$ such that $|a_m - L| < \epsilon$ and $|a_n - L| < \epsilon$. I am not sure how to reconcile these inequalities to get (2) though.

Answer 1

Let $\{a_n\}$ be a non-Cauchy sequence. Then there exist a (fixed) $\varepsilon \gt 0$ so that for any $N$ we can find $s \ge N$ and $t \ge N$ such that $|a_s - a_t| \ge \varepsilon$. We can certainly take $s$ and $t$ so that $a_s \le a_t$.

We can actually construct two subsequences of $\{a_n\}$ by taking the 'next' $N$ to be larger than all the $s ^{\;'}$s and $t ^{\;'}$s seen so far, with the two subsequences satisfying

$\tag 1 \{a_{s_k}\}_{\, k \ge 0} \text{ and } \{a_{t_k}\}_{\, k \ge 0} \text{ with } a_{s_k} + \varepsilon \le a_{t_k}$

Suppose now that $\{a_n\}$ converges to $L$. Then every subsequence converges to $L$. But if we apply the limit operation to $(1)$, we get

$\qquad L + \varepsilon \le \lim\inf\, a_{t_k} = L = \lim\sup\, a_{t_k}$

which is absurd.

Answer 2

Let $\lim_{n\to \infty}a_n=L.$ For $\epsilon >0$ let $\epsilon'=\epsilon /2$ and let $n_{\epsilon'}\in \Bbb N$ such that $$\forall n\geq n_{\epsilon'}\;(|L-a_n|<\epsilon').$$ Then for all $m,n$ such that $n_{\epsilon'}\leq n\leq m$ we have $$|a_n-a_m|=|(a_n-L)+(L-a_m)|\leq |a_n-L|+|L-a_m|<\epsilon' +\epsilon'=\epsilon.$$