1
$\begingroup$

If we say $\{b_n\}$ is a sequence of positive numbers s.t $b_n \to 0$ and if $\{a_n\}$ satisfy $|a_m-a_n|\le b_n$ but we should take $m\ge n$ (for all), how can we say $a_n$ is a Cauchy sequence?

$\endgroup$
2
$\begingroup$

Given $\epsilon \gt0 $ there is $n_0$ s.t for each $n \ge n_0 $ $b_n \lt \epsilon $

Now, for each $n,m \ge n_0 $ it is true that

$|a_n - a_m | \le b_n \lt \epsilon$ , so $\{a_n\} $ is a Cauchy sequence.

$\endgroup$
0
$\begingroup$

The sequence $\{a_n\}$ is Cauchy if for all $\varepsilon>0$ there exists an $N$ such that for all $m,n\geq N$ we have $|a_m-a_n|<\varepsilon.$

So, let $\varepsilon>0$. By assumption there exists $N$ so that for all $n\geq N$, we have $0<b_n< \varepsilon$. Then, for all $m,n\geq N$ we have $|a_n-a_m|<b_n<\varepsilon$, so the sequence is Cauchy.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.