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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. – Brian M. Scott Nov 18 '12 at 22:01
Could we say that $\exist N_1$ s.t. $|a_n - L| < \epsilon / n$? whenever $n > N_1$? – Hawk Nov 18 '12 at 23:13
No: you’re using $n$ to mean two different things on the two sides of the inequality. – Brian M. Scott Nov 18 '12 at 23:28
What do you mean? I am using the same 'n' here. That's the goal of my question (the original one) – Hawk Nov 18 '12 at 23:35
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. – Brian M. Scott Nov 18 '12 at 23:40
up vote 2 down vote accepted

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 $n > N_{\epsilon}$ – Hawk Nov 14 '12 at 3:43
@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? – Brian M. Scott Nov 14 '12 at 3:47
Do you mind coming back to this? Is epsilon here fixed? – Hawk Dec 20 '12 at 20:53
@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 $\epsilon$. – Brian M. Scott Dec 21 '12 at 0:39
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? – Hawk Dec 21 '12 at 1:11

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