# Proving a sequence is Cauchy, given that $|x_n − x_m|< 1/\min\{n, m\}$

Let $(x_n)$ be a sequence of real numbers with the property that for all positive integers $n$ and $m$, $|x_n − x_m|< 1/k$ where $k = \min\{n, m\}$. ￼Prove that the sequence $(x_n)$ is a Cauchy sequence

This is what I have so far:

The definition of a Cauchy sequence is, if given $\epsilon > 0$, there exists and $N$ such that for all $n, m> N$, $|x_n-x_m|< \epsilon$. This is what I have so far for my proof:

Let $\epsilon >0$ be given Pick an $N > 1/\epsilon$, such an $N$ exists by the Archimedean Property of $\mathbb{R}$. This choice of $N$, for any $n,m> N$ we have that $|x_n-x_m|<1/k < 1/n <\epsilon$. I don't fully understand what $k=\min\{n,m\}$ means and how to incorporate this into my proof.

• Please show what you tried on this problem. – user84413 Feb 25 '16 at 19:16
• It should be a fairly simple application of the definition. Just pick $\epsilon>0$, and show that there's $N\in\mathbb N$ such that$|x_n-x_m| < \epsilon$ for $n,m\geq N$. Hint: use the fact that there is a natural number $k$ such that $1/k < \epsilon$. – Trevor Norton Feb 25 '16 at 19:25
• Well, first define what the definition of a cauchy sequence is. Then demonstrate that this sequence satisfies us. Then write us and tell us why and where you had trouble doing this. Okay, a hint. If m is a large integer 1/m is a small fraction. That should be enough if you know what the definition of a cauchy sequence is. – fleablood Feb 25 '16 at 19:29
• Welcome to Stackexchange. Simple "Here's my exercise, solve it for me" posts usually get few responses. It is better to say what you understand about the problem, what you've tried so far, etc. Something to both show you are part of the learning experience and to help us guide you to the appropriate help. You can check this link for more tips. – Frentos Feb 25 '16 at 19:32
• The definition of a cauchy sequence is if given epsilon > 0, there exists and N such that for all n, m> N |xn-xm|< epsilon. This is what I have so far for my proof: Let epsilon >0 be given Pick and N > 1/epsilon, such an N exsits by the Archimedean Property of R This choice of N, for any n,m> N we have that |xn-xm|<1/k < 1/n <epsilon I don't fully understand what k=min{n,m} means and how to incorporate this into my proof – Harry Feb 25 '16 at 20:15

If you note that for all $\epsilon>0$ that there exists an $N \in \mathbb{N}$ so that $\frac{1}{N}<\epsilon$ then it follows for all $m,n \geq N$ that $|x_n-x_m|<\frac{1}{\min(m,n)} \leq \frac{1}{N} < \epsilon$

• I have let epsilon >0 be given and that there exists an N so that 1/N < epsilon and such N exists by the Archimedean Property of R. For any m,n > N that |xn-xm|<1/k<1/N<epsilon. I am unsure how to incorporate that k=min{n,m} – Harry Feb 25 '16 at 19:33
• @Amanda, since $m,n>N$, you know that the minimum of the two is also greater than $N$. In other words: $m,n>N \implies k= \textrm{min}\{m,n\}>N$. The result follows. You're proof is already done, more or less – Andres Mejia Feb 25 '16 at 20:34
• @andresMeijia Great, thank you! I was over-thinking it – Harry Feb 25 '16 at 20:42