Am I allowed to pick ANY $\epsilon$ For negation of a Cauchy sequence? I struggle a bit to understand this concept. Define the sequence ${x}_n$ as follows:
$$ 
{x}_n = \frac{1}{3}, \frac{1}{7}, \frac{1}{11},…,\frac{1}{4n-1}
$$
Use the negation of the definition of Cauchy sequence to show that ${x}_n$ is not a Cauchy sequence.
Negation of Cauchy:
$\exists \space \epsilon \space \forall \space N \space \exists \space n\ge m\gt N \space |{x}_n -{x}_m | \ge \epsilon$
Okay, so I need to find an $\epsilon$. Can I pick ANY epsilon? For example.
Pick n = N+2 and m = N+1, then
$|{x}_n -{x}_m| = |\frac{-4}{(4N+7)(4N+3)}| $
Can I just choose $\epsilon = \frac{3}{(4N+7)(4N+3)}$ ? This doesn’t feel like a legitimate way of choosing $\epsilon$.
 A: *

*This is a Cauchy sequence, so something is wrong if you're trying to prove it isn't.


*You can pick $\epsilon$ when you are trying to disprove that a sequence is Cauchy, but you have to pick it before $N$ gets picked. So it's not allowed to make $\epsilon$ a function of $N$, as you have done. You can see this by reading your quantifiers in order. You have to prove that there exists epsilon such that for all N, ...
A: This sequence clearly converges to $0$ so it's Cauchy.
You can't quite just pick epsilon.   You made a mistake.   If it weren't Cauchy, you could show it by finding epsilon for which the Cauchy criterion isn't met.
Let $x_n=n$.  This sequence isn't Cauchy.  I can produce $\epsilon $ such that for any $N$, there are $n,m\gt N$, such that $\lvert x_n-x_m\rvert \gt\epsilon $.  Try it.

Less trivially,  let $x_n=1+\dfrac 12+\dots +\dfrac 1n$, the so-called harmonic series,  provides a sequence (of partial sums) that is not Cauchy.
One can show that there's $\epsilon $ such that for any $N$, there exist $n,m\gt N$ with $\lvert x_n-x_m\rvert \gt\epsilon $.
It turns out (and this one is tricky), that epsilon can be chosen to be $\dfrac 12$.
To see it, we choose $n\gt2m\gt N$, and look at $$\lvert x_n-x_m\rvert \ge\dfrac 1{2m}+\dots +\dfrac 1{m+1}\ge m\cdot \dfrac 1{2m}=\dfrac 12$$.
