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Okay so we all know the epsilon-N argument for convergence of sequences, that is a sequence $a_n$ converges to $a$ if $\forall \epsilon > 0, \exists N \in \mathbb{N} : n > N \implies |a_n - a| < \epsilon$

Now some point in my life, I've been told that any $\epsilon$ works, but I just cannot choose an $\epsilon$ that is dependent on $n$ because we would get a "changing epsilon".

So for instance, in proving the sum law for limits $\lim_{n\to \infty} a_n +b_n = L +M$ (provided the individual sequences' limits exists) we choose $\epsilon$ to be $\epsilon/2$ for for the partial sequences. But what happens if we choose $\epsilon/n$? So

$|a_n + b_n- L - M| \leq |a_n - L| + |b_n-M| < \frac{\epsilon}{2n}+ \frac{\epsilon}{2n} = \frac{\epsilon}{n}$. Okay so clearly $n$ is still positive, and I kinda see why writing $\epsilon$ in terms of $n$ here is dangerous, but when $n$ is big, can't still say $\epsilon/n < \epsilon$?

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    Yes, if $n>1$, then $\epsilon/n<\epsilon$. But I think that you’re still misunderstanding what’s going on, since (1) you’re using $n$ to mean two completely different things, and (2) you completely neglect the really important part, which is showing that a suitable $N$ exists. We *could* say that there is an $N_1$ such that $|a_n-L|<\epsilon/3$ when $n>N_1$, and there is an $N_2$ such that $|b_n-M|<\epsilon/3$ when $n>N_2$, and we could then conclude that $|(a_n+b_n)-(L+M)|<\epsilon/3+\epsilon/3=2\epsilon/3<\epsilon$ whenever $n>\max\{N_1,N_2\}$, but there’s no good reason to do so.2012-11-18
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    Could we say that $\exist N_1$ s.t. $|a_n - L| < \epsilon / n$? whenever $n > N_1$?2012-11-18
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    No: you’re using $n$ to mean two different things on the two sides of the inequality.2012-11-18
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    What do you mean? I am using the same 'n' here. That's the goal of my question (the original one)2012-11-18
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    You mean that the $n$ of $\epsilon/n$ is the same as the $n$ of $a_n$, so that (for instance) $|a_{100}-L|<\epsilon/100$, while $|a_{200}-L|<\epsilon/200$? Even if $\lim_{n\to\infty}a_n=L$ there’s no guarantee that such an $N_1$ exists, and the definition of convergence doesn’t say that it does.2012-11-18
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    *You mean that the n of ϵ/n is the same as the n of an, so that (for instance) $|a_{100}−L|<ϵ/100$, while $|a_{200}−L|<ϵ/200$?* Yes that's exactly what I mean. And I've been told that I can'd do that. If no $N_1$ exists, doesn't that mean the sum law fails since supposedly, *every* epsilon works? I know this question sounds totally absurd and stupid2012-11-19

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It isn’t $\epsilon$ that ‘works’ or fails to do so: it’s $N$. Look at that definition again: for every possible choice of positive $\epsilon$ there must be an $N_\epsilon$ such that (something nice) happens. In order to prove convergence of the sequence, you must show that no matter what $\epsilon>0$ is given to you, you can demonstrate the existence of an $N_\epsilon$ that ‘works’. Specifically, you must come up with an $N_\epsilon$ so large that $|a_n-a|<\epsilon$ for all $n\ge N_\epsilon$.

Note that it’s $N_\epsilon$ that you produce, not $\epsilon$: you’re given an $\epsilon$, and you have to produce an $N_\epsilon$ that ‘works’ for that $\epsilon$.

As my notation $-$ $N_\epsilon$ instead of just $N$ $-$ should suggest, your $N_\epsilon$ certainly can depend on $\epsilon$. Indeed, in general it must: the closer to $a$ you want to force the terms of the sequence, the further out you’re going to have to go. (In most cases; constant sequences are obviously an exception.)

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    This is going to sound like I didn't read your post at all, but does $\epsilon = \frac{1}{n}$ work? Since $N_{epsilon}$ is "sol large" and $n > N_{\epsilon}$2012-11-14
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    @sizz: The question is meaningless as stated, I’m afraid. As I said, it isn’t $\epsilon$ that works or doesn’t work, at least as the expression ‘works’ is usually used in this context. What exactly do *you* mean by ‘works’ here?2012-11-14
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    Do you mind coming back to this? Is epsilon here **fixed**?2012-12-20
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    @sizz: Yes and no. No, because to prove that the sequence converges, you must show that something is true for every possible choice of $\epsilon>0$, i.e., for every positive real number. Yes, in the sense that in order to do that, you imagine that you’ve chosen some particular $\epsilon$ (without actually saying what it is) and then show that there is a suitable $N_\epsilon$ to go with that choice of $\epsilon$.2012-12-21
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    So I was implying (2), that *given* an epsilon (which after this point, we fix), then we choose the $N_{\epsilon}$. In one of the comments you said that if I had been *given* an epsilon $\frac{1}{n}$, there is no guarantee that a corresponding $N_{\epsilon}$ exists. OKay so first of all I immediately see what the problem is (correct me here if i am wrong), but since I chose $\epsilon = 1/n$, that's no longer "fixed" and that's why you say there is concern of whether n > N can even happen?2012-12-21
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    @sizz: If by $n$ you mean the index of the sequence, then you’re not talking about a fixed quantity, and it’s flat-out meaningless to (try to) let $\epsilon=\frac1n$. The index of the sequence is a dummy variable, a way to label the terms of the sequence.2012-12-21
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    right I am trying to get to the crux of my problem here now, I think I am closed. It was meaningless because for those reasons I stated above (yours to begin with) right?2012-12-21
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    @sizz: Probably; there’s just enough of a language barrier that I can’t *quite* tell from what you wrote whether you fully understand why it doesn’t make sense.2012-12-21
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    For any epsilon you choose and once you have chosen that epsilon, you stick with it and aren't change it to something else, that's what I mean by **fixed**. So choosing $\epsilon = 1/n > 0$ doesn't make sense because I am not choosing **an** epsilon, I am choosing **many** epsilons and that's why it is **not fixed**2012-12-23
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    @sizz: Okay, that sounds as if you have the right idea now.2012-12-23