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, $\forall \varepsilon>0, \exists N \in \mathbb{N}\ s.t \ \lvert a_n-a_m\rvert< \varepsilon, \forall, m\geq n \geq N$.
 A: 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 $\;m\to\infty\;$"
A: 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.
A: The third step is not that clear to me. However, you could do the following:
Setting $U_N\stackrel{\rm def}{=}\sum_{n=1}^N \lvert a_n - a_{n+1}\rvert$, $(U_N)_N$ is real-valued and convergent, and therefore Cauchy. Let $\varepsilon > 0$ be arbitrary: there exists $N$ such that, for all $m > N$ and $p\geq 0$,
$$
\lvert U_m - U_{m+p} \rvert \leq \varepsilon
$$
Rewriting, this means $\sum_{n=m}^{m+p} \lvert a_n - a_{n+1}\rvert \leq \varepsilon$. By the triangle inequality,
$$
\lvert a_m - a_{m+p} \rvert = \lvert \sum_{n=m}^{m+p} a_n - a_{n+1}\rvert \leq \sum_{n=m}^{m+p}\lvert a_n - a_{n+1}\rvert \leq \varepsilon
$$
for all $m > N$ and $p\geq 0$.
A: Apart from the remarks in the comments, you should simply notice that 
$$
\sum_{n=1}^\infty |a_n-a_{n+1}|<\infty
$$
implies
$$
\sum_{n=1}^\infty a_n-a_{n+1}<\infty,
$$
because absolute convergence implies convergence. But this means that
$$
a_1-a_2+a_2-a_3+a_3-a_4+\ldots
$$
converges, and hence $\{a_1-a_n\}_n$ is a convergent sequence.
