If a sequence converges, then it is Cauchy?

Question

Considering the following proof and its converse: If a sequence converges, then it is Cauchy. That is, if $\lim_{n\to \infty}a_{n} = L$, then given $m>N$, we have that $|a_{m}-a_{n}| < \epsilon$

I tried using the $\epsilon-N$ definition for $\lim_{n\to \infty}a_{n} = L$ and that if something is less than $\epsilon$, then it must be less than $\left.\frac\epsilon2\right.\rbrace$

I am also stuck proving the converse: If a sequence is Cauchy, then it converges.

Hints, please?

Answer 1

Suppose $(x_n)$ converges, say $x_n \to L $. Let $\epsilon > 0 $ be given. By definition, we can find $N \in \mathbb{N}$ such that if $n \geq N$, then

$$ | x_n - L | < \frac{ \epsilon}{2} $$

Similarly, take $M \in \mathbb{N}$ such that if $m \geq M$, then

$$ |x_m - L | < \frac{ \epsilon }{2} $$

Let $K = \max\{ N,M \} $. Then, for all $n,m \geq K$, we have (using triangle inequality)

$$ |x_n - x_m | \leq |x_n - L | + |L - x_m| < \frac{ \epsilon}{2} + \frac{ \epsilon}{2} = \epsilon $$

Hence, by definition, $(x_n)$ is a cauchy sequence.