Prove that $\{a_n\}$ is a Cauchy sequence

Question

$\{a_n\}$ 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.

Answer 1

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 $n>m$.

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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.

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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.

Answer 2

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.

Answer 3

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 $|a_n-a_m|=|a_n-a_{\bar{q}}+a_{\bar{q}}-a_m|\leq |a_n-a_{\bar{q}}|+|a_m-a_{\bar{q}}|\leq \epsilon/2+\epsilon/2=\epsilon$.