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, $\forall \varepsilon>0, \exists N \in \mathbb{N}\ s.t \ \lvert a_n-a_m\rvert< \varepsilon, \forall, m\geq n \geq N$.

I'd go like this:
Assuming $\;m>n\;$ , we have
$$|a_m-a_n|=|a_m-a_{m-1}+a_{m-1}-a_{m-2}+a_{m-2}-a_{m-3}+\ldots+a_{n+1}-a_n|\le$$
$$\le\sum_{k=0}^{m-n-1}|a_{m-k}-a_{m-k-1}|\xrightarrow[m,n\to\infty]{}0$$
The last limit is not actually a double one, but rather "make $\;n\to \infty\;$ and thus also $\;m\to\infty\;$"
