cauchy sequence prove for sequence of positive numbers

Question

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?

Answer 1

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.

Answer 2

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.