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 $|a_n-a_m|<\epsilon, \forall n,m>N.$

Please help!!


2 Answers 2


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$$\lvert a_m-a_n\lvert<b_r<\varepsilon,$$since $r\geqslant p$.

  • $\begingroup$ Isn't the non-negativity of each $b_n$ implied by point 2? $\endgroup$
    – Tikak
    Mar 23, 2020 at 13:01
  • 1
    $\begingroup$ Yes, but I wanted to make explicit that assumption. $\endgroup$ Mar 23, 2020 at 13:05

Let $\epsilon>0$ given.

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

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

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

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


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .