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I am learning about sequences and for example:

A series $\sum_{k = m}^ns_k$ is convergent iff for every $\epsilon > 0[\exists N : m, n > N \Longrightarrow |\sum_{k = m}^ns_k| < \epsilon]$.

Now in this example, if it wasn't for the Archimedan property then this wouldn't hold true - there would be an infinitesimal/infinite element such that any sum is greater than said infinite(simal) element.

In fact, if it wasn't for the property, most of real analysis would fail - the definition of $\epsilon$ involves $\exists n : \frac{1}{n} < \epsilon$.

Does this also have to do with the density of $\mathbb{Q}$ in $\Bbb{R}$? Topologically perhaps?

Don Larynx
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    I think that the fact that the real numbers are a field is important. The fact that they can be given an order that is compatible with the field is important. The fact that every set of real numbers that has an upper bound has a least upper bound is important. In fact these three properties essentially determines the real numbers. – Baby Dragon Nov 24 '13 at 09:18
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    It is hard to say which is "the most important property", but if I had to pick one, it would be the least upper bound axiom. – Prahlad Vaidyanathan Nov 24 '13 at 09:23
  • Ah, and this is a consequence of said axiom @PrahladVaidyanathan – Don Larynx Nov 24 '13 at 09:24
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    An ordered field with the least upper bound property is automatically Archimedean. In fact, $\Bbb R$ is the unique ordered field with the least upper bound property, though this is a bit harder to prove. See this answer and the comments that follow it. – Brian M. Scott Nov 24 '13 at 21:10

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I would say "no". $\mathbb Q$ is Archimedean, but would be terrible for analysis. In contrast, things like the Levi-Civita field are non-Archimedean, but their non-Archimedean analysis is still studied.

What does have to do with the density of $\mathbb Q$ in $\mathbb R$ is the fact that real epsilons are unnecessary: As you point out, for all positive $\varepsilon$, you can find a positive rational number smaller than it. That flavor of statement doesn't really require the Archimedean property, though, as it's also true for instances of the Hyperreals.

As Brian M. Scott said and linked, the least upper bound property leads to the Archimedean property almost immediately, and less obviously, it leads to the field being isomorphic to the reals, which you can read a proof of in a textbook like Spivak's Calculus or in the undergrad thesis Pete L. Clark linked. Since you asked about "real analysis", and the least upper bound property is one of many equivalent defining properties of the reals (taken as given that it's an ordered field, which is certainly not always the case in analysis), I'd have to agree with Prahlad Vaidyanathan that that could be the "most important" property.

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Mark S.
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    Second Paragraph But it does require the Archimedan property, or the rationals are bounded. :) – Don Larynx Nov 29 '13 at 23:54
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    @DonLarynx I was too vague with my "flavor of statement" remark. If there's always a positive rational (ordered fields always have a uniquely isomorphic copy of the rationals as a subfield) less than an arbitrary positive number, then yes you need the Archimedean property. What I was getting at was "For any positive hyperreal, you can find a positive hyperrational smaller than it." and even though there are non-standard hyperrationals, this statement still "has the same flavor" to me because of the good analogy between hyperrationals and rationals, etc. – Mark S. Nov 30 '13 at 02:12