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We have some sequence $(a_{n})$ such that, $\lim_{N \rightarrow \infty}\sum_{n=1}^{N} |a_{n} - a_{n+1}| = L$, then $(a_{n})$ is cauchy.

Here is the begining of an attempt

So given the sum has a limit, we can say:

$\forall \epsilon >0 \space \exists \phi \space natural \space s.t. \space \forall n>\phi, |\sum_{n=1}^{N} |a_{n} - a_{n+1}| - L| < \epsilon$
This allows us to know that $lim_{n \rightarrow \infty}|a_{n} - a_{n+1}| = 0$
Therefore $0 \le|a_1 - a_N|=|\sum_{n=1}^{N} (a_{n} - a_{n+1})| \le \sum_{n=1}^{N} |a_{n} - a_{n+1}| < \epsilon +L$

However I do not quite see how to proceed, any hints?

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    Instead of considering $\sum_{n=1}^N \lvert a_n- a_{n+1}\rvert$, consider $\sum_{n=N}^{N+M} \lvert a_n- a_{n+1}\rvert$. Note that since the series $\sum_{n=1}^\infty \lvert a_n- a_{n+1}\rvert$ is convergent, you have $\sum_{n=N}^\infty \lvert a_n- a_{n+1}\rvert \xrightarrow[N\to\infty]{} 0$. – 2017-02-06
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    That makes sense! I had not thought of working with the tail of the sequence. – 2017-02-06
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    Intuitively, that's the thing to do because "that's what Cauchyness is all about." – 2017-02-06

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Given $p,q$ with $q>p$ we have $$ |a_p-a_q|\le \sum_{i=p}^{q-1}|a_i-a_{i+1}|=\sum_{i=1}^{q-1}|a_i-a_{i+1}|-\sum_{i=1}^{p-1}|a_i-a_{i+1}| $$ Let $\epsilon>0$. Since $$ \lim_{N\to0}\sum_{i=1}^N|a_i-a_{i+1}|=L, $$ there is some positive integer $N_0$ such that $$ |\sum_{i=1}^N|a_i-a_{i+1}|-L|\le \frac{\epsilon}{2} \quad \forall N\ge N_0. $$ Hence, for $p,q>N_0$ we have $$ |a_p-a_q|\le (\sum_{i=1}^{q-1}|a_i-a_{i+1}|-L)+(L-\sum_{i=1}^{p-1}|a_i-a_{i+1}|) \le \frac{\epsilon}{2}+ \frac{\epsilon}{2}=\epsilon $$

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    @Clement C. 's hint was a nice dinner, now this is all you can eat. However, in your last line, is the argument not only true with absolute values and not brackets? – 2017-02-06
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    It is true because $x\le |x|$ – 2017-02-07