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

I was looking at the post:

Cauchy Sequence that Does Not Converge

$a_n = \left(1+\frac{1}{n}\right)^n$. 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...

• I think you're missing part of your equation. Nov 7 '13 at 0:46
• Why do you think anybody is suggesting that it's not Cauchy?
– anon
Nov 7 '13 at 0:57
• Ya I actually mistyped that. I'll rephrase...
– user66360
Nov 7 '13 at 1:00
• I actually the totally wrong question. I meant to ask: why IS the sequence Cauchy, using the definition. Whoops!
– user66360
Nov 7 '13 at 1:03
• Then just edit your question to ask what you meant to ask, rather than keeping an incorrect version of it in place.
– anon
Nov 7 '13 at 1:11

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

• Got it. So if $Q$ is my metric space, with $d(x,y)=\mid x-y \mid$. Then I have a Cauchy sequence that does not converge. Correct?
– user66360
Nov 7 '13 at 0:55
• Yes. you can rephrase this saying that $\mathbb{Q}$ is not complete with respect to the metric $|\cdot|$
– user67133
Nov 7 '13 at 0:57
• But if $a_n$ doesn't converge in $Q$, why is it still Cauchy in $Q$?
– user66360
Nov 7 '13 at 1:06
• For the same reason it is Cauchy in $\Bbb R$: because it satisfies the condition for being Cauchy.
– anon
Nov 7 '13 at 1:11
• I think you're forgetting that the equivalence "Cauchy $\iff$ convergent" only holds in a complete metric space (e.g., $\Bbb R$). Nov 7 '13 at 1:15

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.