# cauchy sequence prove for sequence of positive numbers

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?

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.
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.