Quantifier Order in Cauchy Convergence

Question

I'm getting my quantifiers mixed up. The professor put both statements on the board and asked us to think about which implies convergence.

1. $(\forall \epsilon > 0)(\forall k>0 )(\exists N)$ such that ($\forall n \geq N$) $|a_{n+k}-a_n| < \epsilon$

2. $(\forall \epsilon > 0)(\exists N)(\forall k>0)(\forall n \geq N$) $|a_{n+k}-a_n| < \epsilon$

To me they both look like cauchy convergence. I was worried about the placement of $k$, but it doesnt seem to matter because we worry only about $\exists N$ coming before $n\geq N$. The professor mentioned they were different though -- I can't see how.

Answer 1

A suitably chosen slowly growing sequence can satisfy 1.

For instance, consider the sequence $a_n = \log n$ and notice that for any $\epsilon > 0$ and for any $k \geq 1$, we may pick $N$ so that $N > k/\epsilon$. Then for all $n \geq N$ we have

$$ | a_{n+k} - a_n | = \log \left(1 + \frac{k}{n}\right) \leq \frac{k}{n} < \epsilon $$

but of course $a_n \to \infty$ as $n\to\infty$.

Answer 2

It should be the second one. $N$ should depend only on $\epsilon$, and not depend on $k$.

Answer 3

Cauchy (for a sequence of real numbers) should say that for all $\epsilon>0$ there exists an interval of width at most $\epsilon$ such that all but finitely many terms of the sequence are in the interval. This ensures that the terms all eventually get close together.

The first one ultimately implies that the next arbitrarily large number of terms $a_{n+1},\dots,a_{n+k}$ have to be close to $a_n$ once $n \geq N(k)$. (Technically this is not what it says, but it's an easy corollary.) But now the remaining terms are not required to be close to $a_n$. As a result you can, for example, have a sequence that blows up, but at a decreasing rate, so that $|a_{n+k}-a_n| \to 0$ as $n \to \infty$ whenever $k$ is fixed. Lots of functions will do that, including $\log(n),n^a$ for $0<a<1$, etc.