Why is the sequence $ a_n = \left(1+\frac{1}{n}\right)^n $ Cauchy?

Question

I was looking at the post:

Cauchy Sequence that Does Not Converge

And the top answer was this sequence:

$ a_n = \left(1+\frac{1}{n}\right)^n$. I understand that this sequence converges to $e$, which is not a rational number, and that $a_n$ is a sequence of rationals, but I don't see why that proves that $a_n$ it is Cauchy. I wonder if someone could give another explanation of why $a_n$ is Cauchy using the following definition:

A sequence $p_n$ is Cauchy if $\forall \epsilon >0, \exists M \in R$ such that $\forall i, j \in N:$

$i,j > M \implies \mid p_i - p_j \mid < \epsilon$

EDIT: I originally asked the opposite of what I meant to asked. My apologies...

Answer 1

In $\Bbb Q$ this sequence does not converge. In $\Bbb R$ it does. The sequence is Cauchy, regardless.

Answer 2

This sequence converges to $e$. Since it converges, it is necessarily Cauchy.

Suppose $x_n\to l$ as $n\to\infty$. Choose $\epsilon > 0$. Then there is some $N$ so $n\ge N\implies |x_n -l| < \epsilon/2$. Choose $m, n \ge N$. Then $$|x_m - x_n| \le |x_n-l| + |x_m - l| < \epsilon/2 + \epsilon/2 = \epsilon.$$ The sequence is Cauchy.