Definition of convergent, can the N change if I change to another $\epsilon$

Question

I’m having problems with the definition of converging sequence. In my lecture note, the definition is: a real sequence $a_n$ converges to the limit a if for all $\epsilon>0$ there exist a natural number N such that $|a_n-a|<\epsilon$ for all n>N I notice that convergence needs to satisfy for all $\epsilon$, but my question is: if I change my $\epsilon$, can my N change accordingly? Or it has to be no matter how I choose my $\epsilon$, the N after which my sequence is infinitely close to a is always that same N?

I hope I make myself clear. Thank you!

Answer 1

There are already perfectly valid answers. However, I will add something which OP might find useful.

I find thinking about this in terms of agents much more intuitive. Imagine this as a game with two players: you and your opponent. You are tying to prove that the sequence converges to a limit $a$, your opponent would like this to be false. The game is played as follows: Your opponent gives you an $\epsilon$ and you try to find an $N$ such that for all sequence elements after that $N$ the absolute value of the difference between all sequence elements following $N$ and $a$ is smaller than the $\epsilon$ your opponent has given you. Note that this is indeed the definition of convergence.

Practically, it is in your opponent best interest to give you the smallest $\epsilon$ possible; e.g. $0.000000000001$. However, you are always allowed to find a new $N$ such that the absolute value of the difference between all following sequence elements and $a$ is smaller than $\epsilon$. And if you can do this, no matter what $\epsilon$ he gives you, than you have effectively proven that the sequence converges.

Therefore, to answer your question again. Yes, you can indeed change the $N$ if you change the $\epsilon$, that is literally the game your are playing.

Answer 2

Please don't write or think something like

the $N$ after which my sequence is infinitely close to $a$

The whole point of the fancy definition for "convergent" is to avoid saying " "infinitely close".

Think instead that

you can make the terms of the sequence as close as you like to $a$ as long as you look far enough out in the sequence.

How far out you have to look (that's beyond $N$) depends on how close you'd like to be (that's $\epsilon$). So $N$ will depend on $\epsilon$.

Answer 3

Yes, the $N$ is allowed to change. That is why it comes ‘after’ the $\epsilon$ in that sentence. In general, things appearing later on in a sentence depend on the stuff before.

If I wanted to say the $N$ is not allowed to depend on the $\epsilon$, then I should write “there is an $N$ such that for all $\epsilon$…”

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And while we’re talking about meanings of sentences, since we see that the order of description matters a lot, I would strongly suggest you get into the habit of writing all the quantifiers before the formulae. So rather than

“… such that $|a_n-a|<\epsilon$ for all $n>N$”,

write

… such that for all $n>N$, we have $|a_n-a|<\epsilon$

or

such that for all $n$, if $n>N$ then $|a_n-a|<\epsilon$.

In this very simple example, everyone knows the intended meaning regardless of which of these three you write. But, as sentences get longer, you’ll realize that ‘trailing’ quantifiers only serve to confuse. Also, many people write things as in the first example, because that’s how we usually speak in plain English, but still, I’d recommend resisting the temptation.

Answer 4

Generally, your $N$ might be a function of $\varepsilon$, and may change depending on the $\varepsilon$ used. (And this should make sense: if you want $\varepsilon$ to be smaller - and hence all future $a_n$ that much closer to the limiting value $L$ - you will often have to go out much further in the sequence.) Formally, this is because the definition,

"for all $\varepsilon>0$, there exists $N \in \mathbb{N}$ [...]"

begins by defining $N$ after it introduces $\varepsilon$.

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As an example, this past MSE post discusses the example of proving $$ \lim_{n \to \infty} \frac{6n}{n^2+1} = 0 $$ and, concludes that $N > \frac 6 \varepsilon$. (This is a sufficiently narrow definition for $N$, though you could also choose more explicitly, say, $N = \lceil \frac 6 \varepsilon\rceil + 1 $ if it is your preference.)

This definition is dependent on $\varepsilon$, and that's totally fine.

Answer 5

Not only $N$ can change depending on $\epsilon$, the only convergent sequences where $N$ would be independent of $\epsilon$ are eventually constant, i.e. $a_n = a_m$, for all $m,n\geq N$.

To see that this is true, let $(a_n)$ be convergent with limit $a$ and assume that there exists $N$ such that $$(\forall \epsilon >0)\ n\geq N \implies |a_n-a|<\epsilon.\tag{1}$$

Take any $n$ such that $n\geq N$. If $|a_n-a| >0$, let $\epsilon = |a_n-a|$. Then $|a_n-a| \geq \epsilon$, contradicting $(1)$. Therefore, $|a_n-a| = 0$, i.e. $a_n = a$.

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The point is that not only you can change $N$ depending on $\epsilon$, the chances that you won't have to do it given a random convergent sequence are effectively zero, as JonathanZ pointed out in the comments.