Prove that the sequence has a limit I got a problem in my homework I'm having troubles to solve.
Let ${a_n}$ be a sequence such that for every n $\in \mathbb N$
$|a_2 − a_1|+|a_3 − a_2|+...+|a_n− a_{n-1}| < c$ 
$c\in \mathbb R$ 


*

*prove that ${a_n}$ has a limit.


I took ${b_n}$ to be $|a_2 − a_1|+|a_3 − a_2|+...+|a_n− a_{n-1}|$
$0\le{b_n}\lt c$ and ${b_ {n-1}- b_n}\ge 0$, therefore monotone and bounded and therefore is a cauchy sequence. 
So for every $\epsilon \gt 0$ and $P \in \mathbb N$  ,there is such $N \in \mathbb N$ that for every $n>N$,
$|{b_{n+p}- b_n}|< \epsilon$
But I'm stuck at this stage.
I would appreciate if anyone could help me. Thanks in advance.
 A: Since $\sum_{n=1}^N|a_n-a_{n-1}|<c$ for all $N\in \Bbb N$ by hypothesis, the series is convergent if we take the limit as $N\to\infty$. In particular, for any $\varepsilon>0$, we can take $J$ large enough so that $\sum_{n=J}^\infty|a_n-a_{n-1}|<\varepsilon$. Thus, for any $j,k>J$, we have
$$
|a_j-a_k|\leq|a_{j+1}-a_j|+\cdots+|a_k-a_{k-1}|\leq\sum_{n=J}^\infty |a_n-a_{n-1}|<\varepsilon.
$$
Hence, the sequence $\{a_n\}_{n=1}^\infty$ is a Cauchy sequence and is therefore convergent.
A: Rephrasing:
$ S_n := \sum_{k=2}^{n} |a_k-a_{k-1}|;$
$S_n$ bounded above, increasing, hence convergent.
$\rightarrow:$
$S_n$ is a Cauchy sequence:
For $\epsilon \gt 0$ there is a $n_0$ such that for $m\ge n \gt  n_0:$
$|S_m-S_{n-1}| \lt \epsilon$, i. e.
$\sum_{k=n}^{m} |a_k -a_{k-1}| \lt \epsilon$. 
We have:
$ |a_{m} - a_{n-1}| =$
$ |a_{m} - a_{m-1} + a_{m-1} -a_{m-2} +a_{m-2} $
$....+a_n -a_{n-1}| \le $
$|a_{m} -a_{m-1}| +...|a_{n} - a_{n-1}| =$
$|S_m-S_{n-1}|  \lt \epsilon.$
Hence $a_n$ is Cauchy , hence convergent.
