# Determine if a sequence is Cauchy

Let $(x_n)$ be a sequence such that $(\forall m,n \in \mathbb{N}), |x_m-x_n|\ge\frac{1}{\min(m,n)}$. Is $(x_n)$ a Cauchy sequence? I know that a sequence is Cauchy if for every $\epsilon$ there exists $n_0\in \mathbb{N}$ such that for $n,m\ge n_0$ it follows that $|x_n-x_m|<\epsilon$. But here we have $|x_m-x_n|\ge \frac{1}{\min(m,n)}$. How can I prove whether this sequence is indeed a Cauchy. I either need to find $n_0$ or another way I guess is to try to prove that it converges. Frankly I'm not sure I even understand the statement.

• Let $x_n = n$ for all $n$. Then $|x_m-x_n| \ge 1 \ge 1/\min(m,n)$. So we can't say that $(x_n)$ is Cauchy. – GNUSupporter 8964民主女神 地下教會 Mar 1 '18 at 22:23
• Such a sequence cannot exist because for $m = n$ we have $$|x_n - x_n| = 0 \not\geq \frac{1}{n}$$ – mechanodroid Mar 1 '18 at 22:24
• If it's an exercise, especially if to prove $(x_n)$ is Cauchy, then I guess this is a typo only, and they intended to write $\le$ in place of $\ge$. – Berci Mar 1 '18 at 22:26
• So I guess the point was to find an example. Thanks, makes sense now. – DreaDk Mar 1 '18 at 22:29

I think the likelihood is that you've misread the claim, or that there is a typo in the claim.

Firstly, it does not specify that $n \neq m$. If $n = m$ then $|x_{n}-x_{m}|=0$ so $0 \geq \frac{1}{\min{(n,m)}}$ which is a contradiction, so there are no such sequences as written.

If we refine the condition slightly so that $n\neq m$ then note that $\min {(n,m)} \geq 1$ so that $1 \geq \frac{1}{\min{(n,m)}}$ so, as Giuseppe mentioned $x_{n} = n$ works (since $|n-m| \geq 1 \geq \frac{1}{\min{(n,m)}}$).

Now suppose that we take what's probably the correct interpretation of the claim: if $(x_{n})$ is a sequence such that $(\forall n,m \in \mathbb{N}) |x_{m}-x_{n}| \leq \frac{1}{\min{(n,m)}}$, then $(x_{n})$ is Cauchy.

In this case, the strategy you outlined works perfectly well. We fix some $\epsilon > 0$ and look directly for some $n_{0}$ for which we can guarantee that $n,m>n_{0} \implies |x_{n}-x_{m}|<\epsilon$. The simple way to do this is to see that $n,m > n_{0} \implies \frac{1}{\min{(n,m)}} < \frac{1}{n_{0}}$, so all we need is some $n_{0}$ such that $\frac{1}{n_{0}} < \epsilon$; in other words, $n_{0} > \frac{1}{\epsilon}$ we we know must exist by the Archimedian property of $\mathbb{R}$. (For instance, $n_{0} = \lceil \frac{1}{\epsilon} \rceil$ works.) If you're uncomfortable with this last step, it's worth learning more about the Archimedean principle, which is beyond the scope of this question.

• It' a past exam question that I have in front of me. I think it's supposed to be more of a find an example that contradicts it. – DreaDk Mar 1 '18 at 23:08
• @DreaDk if you're looking for a counter-example to the statement as written, do the two I gave you make sense? – Oscar Heath Mar 8 '18 at 1:14

Indeed, the answer is negative. If you consider the sequence $x_n=n+1$, this satisfies the hypotesis clearly but it is not a Cauchy sequence since it diverges.