### Question

A sequence $\{a_n\}$ of real numbers is said to be a Cauchy sequence of for

each $\epsilon$ > 0 there exists a number $N > 0$ such that m, $n > N$ implies

that $|a_n − a_m| <\epsilon$.Prove that every convergent sequence is a Cauchy sequence

### Attempt

This is my first time hearing what a cauchy sequence is. I have no idea how to even start this. I googled cauchy sequence and I think its when $a_n$ converges to $a_{n+1}$?

Attempt:

WTS: $\exists a_m \in \mathbb R, \forall \epsilon > 0, \exists N > 0$, such that for all $n \in \mathbb N$, if $n > N$, then $|a_n – a_m| < \epsilon$

Let $\epsilon > 0$ be arbitrary

Choose N such that for $n > N$ we have $|a_n – a_m| < \epsilon$

Suppose $n > N$, then

??

Could someone point me to the right direction? Thx.

## Best Answer

The sequence $(a_{n})$ is convergent by assumption; let $l := \lim_{n}a_{n}$. Let $\varepsilon > 0$. Then there is some $N$ such that $|a_{n} - l| < \varepsilon/2$ for all $n \geq N$. Note that $$ |a_{n} - a_{m}| \leq |a_{n}-l| + |l-a_{m}| < \varepsilon/2 + \varepsilon/2 = \varepsilon $$ for all $n,m \geq N$.

The major tool here is the triangle inequality.