Cauchy Sequence $|a_n - a_m| < \epsilon $ Question


  A sequence $\{a_n\}$ of real numbers is said to be a Cauchy sequence of for
  each $\epsilon$ > 0 there exists a number $N &gt; 0$ such that m, $n &gt; N$ implies
  that $|a_n − a_m| &lt;\epsilon$.
  
  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: $\exists a_m \in \mathbb R, \forall \epsilon &gt; 0, \exists N &gt; 0$, such that for all $n \in \mathbb N$, if $n &gt; N$, then $|a_n - a_m| &lt; \epsilon$

Let $\epsilon &gt; 0$ be arbitrary

Choose N such that for $n &gt; N$ we have $|a_n - a_m| &lt; \epsilon$

Suppose $n &gt; N$, then 

??

Could someone point me to the right direction? Thx.
The sequence $(a_{n})$ is convergent by assumption; let $l := \lim_{n}a_{n}$. Let $\varepsilon &gt; 0$. Then there is some $N$ such that $|a_{n} - l| &lt; \varepsilon/2$ for all $n \geq N$. Note that 
$$
|a_{n} - a_{m}| \leq |a_{n}-l| + |l-a_{m}| &lt; \varepsilon/2 + \varepsilon/2 = \varepsilon
$$
for all $n,m \geq N$.

The major tool here is the triangle inequality.
If it is not Cauchy, it is obviously not convergent: for |a_n -a_m| is greater than some epsilon for arbitrarily large n,m. This assures a_n and a_m do not both get within epsilon/2 of some a_infinity.