2
$\begingroup$

Let $\{x_n\}$ be a sequence of real numbers with $|x_n − x_m| < 1/(n+m)$ where $n,m$ are positive integers. . k  Prove that the sequence $\{x_n \}$ is Cauchy.

Here is what I have so far..

Let $\epsilon >0$ be given. There exists $N> 1/\epsilon$ by the Archimedean property.

For any $n, m≥N$

$|x_n-x_m|< 1/(n+m) < 1/N < 1/(1/\epsilon)) = \epsilon$

This proves the sequence is Cauchy. I am not sure my proof is correct when I jump from $1/(n+m) <1/N...=\epsilon$.

$\endgroup$
  • 2
    $\begingroup$ That step is correct. If $n,m>N,$ then $n+m \geq 2N > N$, so $\frac{1}{n+m} \leq \frac{1}{2N} < \frac{1}{N}$. The proof is perfect. $\endgroup$ – астон вілла олоф мэллбэрг Mar 25 '16 at 0:18
  • $\begingroup$ Or simply, $n + m > n$, so $1/(n+m) < 1/n \leq 1/N$. $\endgroup$ – Bungo Mar 25 '16 at 1:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.