Am I allowed to pick ANY $\epsilon$ For negation of a Cauchy sequence?

Question

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 $\epsilon$.

Answer 1

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

2. 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, ...

Answer 2

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.

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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 $$\lvert x_n-x_m\rvert \ge\dfrac 1{2m}+\dots +\dfrac 1{m+1}\ge m\cdot \dfrac 1{2m}=\dfrac 12$$.