Prove that $\{a_n\}$ is a Cauchy sequence $\{a_n\}$ and $\{b_n\}$ are sequences and the limit of $b_n$ is $0$. Let $|a_n−a_q| ≤ b_q$ whenever $n,q ∈ N $ and $n ≥ q$. Prove that ${a_n}$ is a Cauchy sequence.
I have already proven that $b_q < \epsilon$ for all $q \geq K$.
I know I need to prove that $ |a_n - a_m| < \epsilon$ by looking at two cases:
when (i) $n\geq m$ and (ii) $n < m$, using $b_q < \epsilon$ for all $q \geq K$ and $|a_n - a_q| \geq b_q$ when $ n \geq q$ but I am stuck on how to do so.
 A: It is not necessary to look at two cases because
$$|a_{n}-a_{m}|=|-(a_{m}-a_{n})|=|a_{m}-a_{n}|.$$
Therefore, we can assume $n>m$.

A small nitpick is that we require $b_{q}\geq0$ for all $q$ since otherwise the inequality $|a_{n}-a_{q}|\leq b_{q}$ can not hold.  Of course, we could argue that by saying the inequality holds we are implicitly making this assumption, but, better safe than sorry.

Now, to the execution. You have mentioned that for any $\varepsilon>0$ we can find a natural number $K$ such that whenever $k>K$ we have
$$|b_{k}| =b_{k}<\varepsilon.$$
Let $m,n>K$ be natural numbers with $m>n$ then by assumption
$$|a_{m}-a_{n}|\leq b_{n}< \varepsilon.$$
Therefore the sequence $(a_{n})$ is Cauchy.
A: Choose $\bar{q}$ such that for $q\geq\bar{q}$ you have $b_q\leq\epsilon/2$, then if $n,m\geq\bar{q}$ you have $|a_n-a_m|=|a_n-a_{\bar{q}}+a_{\bar{q}}-a_m|\leq |a_n-a_{\bar{q}}|+|a_m-a_{\bar{q}}|\leq \epsilon/2+\epsilon/2=\epsilon$.
A: Let $n,m\in \mathbb{N}$ WLOG assume that $n<m$ we should estimate
$$|a_n-a_m|\leq |b_q|$$
By our assumption and as  André Armatowski say and point in their answer.
Since $b_n\to 0$ when $n\to \infty$ then for any $\varepsilon>0$ there are
$N\in \mathbb{N}$ we have $|b_N|\leq \varepsilon $, in particular is is true for $q\in \mathbb{N} $ and $|b_q|<\varepsilon $ hence
$$|a_n-a_m|\leq b_q<\varepsilon $$
Therefore $\lbrace a_n \rbrace$ is a cauchy sequence as well.
