Proving that a sequence is not Cauchy

Question

I'm trying to prove that the sequence $\left(\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},\cdots\right)$ is not a Cauchy sequence.

I know that a sequence of real numbers is not Cauchy if there exists an $\epsilon>0$ such that, for all $N\in\mathbb{N}$, there exists $m,n>N$ such that $|x_{m}-x_{n}|\geq\epsilon$. It intuitively makes sense to me that the sequence cannot be Cauchy, as the distance between points where the denominator changes (like $\cdots\frac{99}{101},\frac{100}{101},\frac{1}{102},\cdots$) keeps growing larger. However, I'm not sure how to find indices $m$ and $n$ in general with $|x_{m}-x_{n}|\geq\epsilon$. Thanks in advance for any help!

Answer 1

Let $(a_n)_{n\in\mathbb N}$ be your sequence. Take $\varepsilon=\frac12$. Given $N\in\mathbb N$, take $n\geqslant N$ such that $a_n$ is of the form $\frac k{k+1}$ for some $k\in\mathbb N$ and let $m=n+1$. Then $a_m=\frac1{k+2}$. Therefore,$$\left\lvert a_m-a_n\right\rvert=\frac k{k+1}-\frac1{k+2}\geqslant\frac12=\varepsilon.$$

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Or you can say that your sequence diverges, since the subsequence$$\frac12,\frac13,\frac14,\ldots$$converges to $0$, whereas the subsequence$$\frac12,\frac23,\frac34,\ldots$$converges to $1$. So, the sequence diverges and therefore it is not a Cauchy sequence.

Answer 2

You only need such an $\epsilon$ to exist, so you can choose a convenient value.

Then you need to show that there are gaps bigger than $\epsilon$ between elements of the sequence as far a=out as you care to go - that gaps as big as $\epsilon$ don't fade out and disappear in the tail of the sequence.

Well you have identified some chunky gaps which persist (you don't need every gap to be big). What $\epsilon$ would work for the gaps you have identified?

Answer 3

Cauchy sequence: $$\forall\varepsilon>0\exists N\forall n,m\quad n,m>N\implies|a_n-a_m|<\varepsilon$$

We want to show the negative:$$\exists\varepsilon>0\forall N\exists n,m\quad n,m>N\land |a_n-a_m|\ge\varepsilon$$

Take $\varepsilon=1/10$, and then, for all $N$ take $n=\sum_{i=1}^{N+1} i=\frac{(N+1)^2+N+1}{2}, m=n+1$ and you have $a_n$ an element in the form of $\frac{k}{k+1}$ and $a_m$ in the form of $\frac{1}{k+2}$