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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...

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

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

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