Let $a_n$ be a sequence such that $\lim_{N\rightarrow \infty} \sum_{n = 1}^{N} |a_n - a_{n+1}| < \infty$. Show that $a_n$ is Cauchy

Question

Let $a_n$ be a sequence such that $\lim_{N\rightarrow \infty} \sum_{n = 1}^{N} |a_n - a_{n+1}| < \infty$. Show that $a_n$ is Cauchy

Attempt:

to show the sequences is Cauchy I have to find $N$ such that for all $m,n \geq N$ one has $|a_m - a_n| < \epsilon$

$$|a_m - a_n| = |a_m - a_{m+1} + a_{m+1} - a_{m+2} + .... - a_{n-1} + a_{n-1} - a_n|$$ $$\leq |a_m - a_{m+1}| + |a_{m+1} - a_{m+2}| + ....+|a_{n-2} - a_{n-1}| + |a_{n-1} - a_n|$$ $$= \sum_{i=m}^{n} |a_i - a_{i+1}|$$

Here is where I am stuck. How can I use the piece I was given to arrive at an Epsilon. Specifically how to use: $\lim_{N\rightarrow \infty} \sum_{n = 1}^{N} |a_n - a_{n+1}| < \infty$

Answer 1

Hint. Let $S_N= \sum_{n = 1}^{N} |a_n - a_{n+1}|$. Then, since $\lim_{N\to \infty}S_N=L\in\mathbb{R}$, it follows that $(S_N)_N$ is a Cauchy sequence. Moreover $\sum_{i=m}^{n} |a_i - a_{i+1}|=S_n-S_{m-1}$.

Answer 2

Convergence of a series means precisely that the tail of the series goes to zero. That's exactly what you need.