# Prove a sequence is Cauchy given information about a second sequence

I've seen a few Cauchy sequence questions, but not this one.

Suppose you have two sequences $$\{a_n\}_{n=1}^\infty$$ and $$\{b_n\}_{n=1}^\infty$$ such that:

1. $$\{b_n\}_{n=1}^\infty$$ converges to $$0$$
2. $$\forall p,q\in\Bbb{Z}^{>0}$$ with $$q \ge p$$, $$|a_q-a_p|\le b_p$$

Prove that $$\{a_n\}_{n=1}^\infty$$ is a Cauchy sequence.

I have no idea how to attempt this question, other than to consider what I need to prove: $$\forall\epsilon>0, \exists N\in\Bbb{N}$$ such that $$|a_n-a_m|<\epsilon, \forall n,m>N.$$

I will assume that each $$b_n$$ is non-negative.

Take $$\varepsilon>0$$. Now, take $$p\in\mathbb N$$ such that $$q\geqslant p\implies b_q<\varepsilon$$. Then, if $$m,n\geqslant p$$, if $$r=\min\{m,n\}$$, we have$$\lvert a_m-a_n\lvertsince $$r\geqslant p$$.

• Isn't the non-negativity of each $b_n$ implied by point 2? Mar 23, 2020 at 13:01
• Yes, but I wanted to make explicit that assumption. Mar 23, 2020 at 13:05

Let $$\epsilon>0$$ given.

$$\lim_{p\to +\infty}b_p=0 \;\; \implies$$

$$\exists N\in \Bbb N \;\; : \forall p\ge N \;\;b_p<\epsilon \implies$$

$$\exists N\in \Bbb N \;\; : \forall q>p\ge N \;\; |a_p-a_q|\le b_p<\epsilon$$

$$\implies (a_n) \text{ is Cauchy}$$