1
$\begingroup$

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.

$\endgroup$
5
  • 3
    $\begingroup$ Please show what you tried on this problem. $\endgroup$
    – user84413
    Feb 25, 2016 at 19:16
  • $\begingroup$ 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 $1/k < \epsilon$. $\endgroup$ Feb 25, 2016 at 19:25
  • $\begingroup$ 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. $\endgroup$
    – fleablood
    Feb 25, 2016 at 19:29
  • $\begingroup$ 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. $\endgroup$
    – Frentos
    Feb 25, 2016 at 19:32
  • $\begingroup$ 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 $\endgroup$
    – Harry
    Feb 25, 2016 at 20:15

1 Answer 1

3
$\begingroup$

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 $|x_n-x_m|<\frac{1}{\min(m,n)} \leq \frac{1}{N} < \epsilon$

$\endgroup$
3
  • $\begingroup$ 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} $\endgroup$
    – Harry
    Feb 25, 2016 at 19:33
  • $\begingroup$ @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 $\endgroup$ Feb 25, 2016 at 20:34
  • $\begingroup$ @andresMeijia Great, thank you! I was over-thinking it $\endgroup$
    – Harry
    Feb 25, 2016 at 20:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .