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I'm curious whether it can be proven, given a convergent sequence {S$_n$}, that some multiple of the limit of this sequence exists. It seems like a pretty simple statement and I'm sure its possible to prove, but I'm somewhat uncertain how to present a rigorous argument. I'm attempting to use the ε, N definition of a limit, but I'm uncertain if what I'm doing, is sufficiently rigorous to provide a legitimate proof.

$lim_{x \to +\infty} S_N=L$ exists ε, N definition: $S_n -> L $ as $n->\infty$ provided that for every number $ε>0$ there is an integer $N$ so that $|S_n - L|<ε$ whenever $n>=N$

So, I simply multiplied everything by a constant and kind of assumed that because everything increases in the same way, that a multiple of the original existing limit will exist. This is not so rigorous though, so I'm looking for a way to make the final leap to a rigorous proof.

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    $\begingroup$ I believe there is something deeper here than intended. What do you mean by exist? Let $\{S_n\}$ have indices in the set $A$. Then do you mean that multiples of the limit exist in $A$? This is not true in general. $\endgroup$
    – Eoin
    Commented Sep 16, 2014 at 1:57
  • $\begingroup$ Generally, as I have seen it in textbooks I've read on limits. Limits are stated as existing and I assume that what is meant by this, is simply that there exists some number L, to which the sequence converges. This is the how I am using the term above. I don't mean in a particular set, although perhaps the way I am using it doesn't actually make any sense. $\endgroup$
    – TQM
    Commented Sep 16, 2014 at 2:19
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    $\begingroup$ You're right. Perhaps I am being pedantic. However, I would just like to clarify as in the limit of a sequence of rationals may exist as an element of the real numbers. If this is what you mean then of course we may achieve multiples. Simply multiply all terms of this sequence by $c$ a scalar and we will achieve $\lim \{S_n\} = L$ and $\lim c\{S_n\} =c\lim \{S_n\} = cL$. However, I am unsure of the extension to other metric spaces. $\endgroup$
    – Eoin
    Commented Sep 16, 2014 at 2:23
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    $\begingroup$ It certainly depends. I believe there is a much more rigorous proof of this. Also if the requirement is as such: given a sequence $\{S_n\}$ with $s_i$ elements of $\mathbb{Q}$ that converges to real number $L$, does there exist a sequence $\{T_n\}$ with $t_i$ in $\mathbb{Q}$, then this is not a proof at all. Because $c$ may not be a rational. However, I believe there is a proof of such a theorem with looser requirements. I will find you a version. $\endgroup$
    – Eoin
    Commented Sep 16, 2014 at 2:54
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    $\begingroup$ here is a set of interesting proofs that you may or may not consider rigorous. I believe that when finding such a statement in texts most of the requirements are implicit. This being said, I would consider a proof to be rigorous if all assumptions are made to be explicit. As for anything else, as long as one step implies the other I'm pretty much satisfied. $\endgroup$
    – Eoin
    Commented Sep 16, 2014 at 3:01

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