# If the sum of the differences of consecutive terms of a sequence between any two terms is less than or equal to one, then the sequence converges

I recently decided to learn real analysis, and started using Krantz's analysis text. I was working through the second chapter on sequences and series and came across this question:

Let $$\{a_j\}$$ be a sequence. Suppose for all $$N>M>0$$, $$|a_M-a_{M+1}|+|a_{M+1}-a_{M+2}|+\ldots+|a_{N-1}-a_N|\leq 1$$. Prove that $$\{a_j\}$$ converges.

First of all, I know that this question is a duplicate of Proof that if the sum of the differences of consecutive terms of a sequence between any two terms is bounded above by one, then the sequence converges. However, I don't quite understand the accepted answer in that question. I want to prove this using only the definition of convergence and Cauchy sequence, either by directly choosing an $$N>0$$ for an arbitrary $$\epsilon>0$$, or through a contradiction, as these are the only thing I've learned so far. When I try to prove by contradiction it just feels like I'm running around in circles, and when I try to prove it directly I haven't the faintest idea how to choose $$N$$. Any help would be greatly appreciated.

Also, I know I'm not suppose to open a new question when there's already a duplicate. If there's something else I could have done, please do inform me in the comment section. Thank you.

Let $$S_n=\sum_{k=1}^n|a_k-a_{k+1}|$$ for $$n\in\mathbb N$$. By hypothesis, $$S_n\leq 1$$ and $$S_n$$ is increasing, so it is convergent.
Given any $$\epsilon>0$$, there exists $$N>0$$ such that for any $$n>m>N$$, we have that $$S_n-S_m<\epsilon$$, which means $$|a_{m+1}-a_{m+2}|+\cdots+|a_{n}-a_{n+1}|<\epsilon.$$ By triangle inequality, we have that $$|a_{n+1}-a_{m+1}|<\epsilon$$, so $$\{a_n\}$$ is a Cauchy sequence, as desired.