Show a sequence such that $\lim_{\ N \to \infty} \sum_{n=1}^{N} \lvert a_n-a_{n+1}\rvert< \infty$, is Cauchy 
Attempt. Rewriting this we have, $$\sum_{n=1}^{\infty} \lvert a_n-a_{n+1}\rvert<  \infty
\,\,\,\Longrightarrow\,\,\, \exists N \in \mathbb{N}\ \ s.t,\  \ \sum_{n \geq N}^{\infty} \lvert a_n-a_{n+1}\rvert < \infty$$
Taking $m>n+1$ we have, $$\sum_{m > n \geq N}^{\infty}\lvert a_n-a_m\rvert< \infty$$
From the above statement we have: $\,\lim_{\ n \to \infty} \lvert a_n-a_m\rvert=0$.
Now combining out statement we have, $\forall \varepsilon>0, \exists N \in \mathbb{N}\ s.t \ \lvert a_n-a_m\rvert< \varepsilon, \forall, m\geq n \geq N$.

If the series $\,\,\sum_{n=1}^\infty\lvert a_{n+1}-a_n\rvert\,\,$ converges, 
then for every $\varepsilon>0$, there exists an $N$, such that
$$
\sum_{n=N}^\infty\lvert a_{n+1}-a_n\rvert<\varepsilon.
$$
Therefore, if $m\ge n\ge N$, then
$$
\lvert a_m-a_n\rvert\le\lvert a_{n+1}-a_n\rvert+\lvert a_{n+2}-a_{n+1}\rvert+
\cdots+\lvert a_{m}-a_{m-1}\rvert\\=\sum_{k=n}^{m-1}\lvert a_{k+1}-a_k\rvert
\le \sum_{k=N}^{\infty}\lvert a_{k+1}-a_k\rvert<\varepsilon,
$$ 
and hence the sequence $\{a_n\}_{n\in\mathbb N}$ is a Cauchy sequence.
