1
$\begingroup$

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 $\epsilon/n < \epsilon$?

  • 0
    *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

1 Answers 1

2

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.)

  • 0
    @sizz: Okay, that sounds as if you have the right idea now.2012-12-23