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I am having a hard time understanding the difference between the two statements,

$$\forall \epsilon >0 \ \exists N \ \in \mathbb{N} \ \forall n\ge N \ (|a_n-a| < \epsilon)$$ and

$$\exists N \in \mathbb{N} \ \forall \epsilon > 0 \ \forall n\ge N \ (|a_n-a| < \epsilon) $$

I can see the difference when I am trying to prove a limit (unless the sequence is constant) because I need to let $N$ depend on $\epsilon$. But when I say the first statement out loud, it seems like it explains what a convergent sequence is doing.

If there is a $N$ s.t. for any positive number you give me, any index greater than that $N$ implies the distance between our limit and the sequence is less than the positive number. What am I missing?

Amzoti
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tmpys
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    The last one means that the sequence is eventually constant: "There is an index $N$ so that any term in the sequence after that is closer to $a$ than any positive bound." – Arthur Oct 24 '14 at 23:04
  • Is it not the case that every convergent sequence eventually becomes constant? – tmpys Oct 24 '14 at 23:08
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    No, it's not. Look at, say, the sequence $a_n=\frac1n$. It converges to $0$ by the Archimedean principle and the (first) definition you've written above, but it never stops moving. – Arthur Oct 24 '14 at 23:14

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Notice how this $N$ is enough for every $\epsilon$ given in this second statement. Let's study a sequence that satisfy this second statement.

Let $(a_n)_{n \in \mathbb{N}}$ be a sequence that satisfies it. Then, there exists $N \in \mathbb{N}$ such that for every $\epsilon > 0$ given, $m>N \implies |a_m - a| < \epsilon$. For every $n \in \mathbb{N}$, you can take $1/n$ for $\epsilon$, thus giving you that $m>N \implies |a_m - a| < 1/n \ \forall n \in \mathbb{N} \implies |a_m - a| = 0 \implies a_m = a$. Then it means that $a_n$ is constant for $m>N$!

Rafael Bordoni
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The second definition says that $N$ does not depend on $\varepsilon$: the same $N$ continues to work now matter how small $\varepsilon$ gets. You want $a_n$ to be within $\varepsilon=0.000000000000000000001$ of $a$? By the second definition, you don't need to make $N$ any bigger than if you want it to be within $\varepsilon=0.001$. That will work if after some finite number of steps, i.e. for $n\ge$ some $N$, we have $a_n$ always equal to $a$. Otherwise then $N$ must depend on $\varepsilon$ and $N$ must get bigger as $\varepsilon$ gets smaller.

Michael Hardy
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  • Is the $2^{\text{nd}}$ statement true if and only if $a_n$ is the constant sequence? – Jack Oct 24 '14 at 23:09
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    @Jack it's true iff $a_n$ is eventually a constant sequence. Very few convergence criteria (none that I know of, at least) really care about what happens to the first terms of $a_n$, for any reaonable use of the word "first" (be it the first five or the first five quadrillion). – Arthur Oct 24 '14 at 23:24
  • Let's say $a_n$ wanders around until $n$ gets to four gazillion, and after that the sequence is constant. In other words, the sequence is "eventually constant". In that case, $N=\text{four gazillion}$ will work no matter how small $\varepsilon$ is. For sequences that are not eventually constant, i.e. constant beyond some value of $N$, the second statement will not hold. (I think people from Germany and France sometimes misuderstand the word "eventually", thinking it means "maybe". If any such are reading this, look the word up.) – Michael Hardy Oct 24 '14 at 23:28