Prove that $(a_1-a_2)+(a_2-a_3)+....$ converges iff ${a_n}$ converges Let $b_n=a_n-a_{n+1}$.
We first assume that ${a_n}$ converges so ,$\lim(b_n)=\lim(a_n)-\lim(a_{n+1})$ , hence $\lim(b_n)=0$.
Now let $s_m$ and $s_n$ be the consecutive partial sums of ${b_n}$.
So $|s_n-s_m| = |a_{n+1}-a_{m}|=|a_{n+1}-L+L-a_m|< \epsilon$ for $N\ge M \ge M(\epsilon)$.So the series is cauchy convergent .
Now let us assume that the series is convergent so the sequence formed by the partial sums are convergent hence $s_M=a_1-a_{M+1}$ hence $-\lim(s_M) +a_1 = \lim(a_{M+1})$.So $A=-S+a_1$.
Can someone go through my attempt and point out my mistake instead of suggesting another answer. 
 A: It's a telescoping series.
$\sum_{k=1}^n (a_k - a_{k+1}) = a_1 - a_{n+1}$.
so $\lim\limits_{n\to \infty} \sum_{k=1}^n (a_k - a_{k-1}) = \lim\limits_{n\to \infty} (a_1 - a_{n+1}) = a_1 - \lim\limits_{n\to \infty} a_n$
Which converges and exists if and only if the $a_n$ converges.
A: 
Let $b_n=a_n-a_{n+1}$.
We first assume that ${a_n}$ converges so
  ,$\lim(b_n)=\lim(a_n)-\lim(a_{n+1})$ , hence $\lim(b_n)=0$.

You've shown that if $b_i$ converges, then it converges to $0$, but you haven't shown that $\lim_{i \to \infty} b_i$ exists. If you're looking to show $b_n \to 0$, then similarly to what you've done for $s_m$ below, you want to argue
$$\lvert b_n \rvert \leq \lvert a_n - a \rvert + \lvert a_{n+1} - a \rvert \leq 2 \varepsilon$$
for all $n > N$ for some $N$, via convergence of $a_n$.
However, you're trying to show that $\sum b_n$ converges, and knowing $b_n \to 0$ doesn't help you here.

Now let $s_m$ and $s_n$ be the consecutive partial sums of ${a_n}$.
So $|s_n-s_m| = |a_{n+1}-a_{m}|=|a_{n+1}-L+L-a_m|< \epsilon$ for $N\ge
> M \ge M(\epsilon)$.So the series is cauchy convergent .

I'm not entirely clear on what you've assumed here, nor exactly what your definition of $s_n$ is. Is it $\sum a_n$ or $\sum (a_n - a_{n+1})$? I think you want the latter, but "consecutive partial sums" means $S_n$ and $S_{n+1}$ for some partial sums $S$, not "partial sum of differences of consecutive terms".

Now let us assume that the series is convergent so the sequence formed
  by the partial sums are convergent hence $s_M=a_1-a_{M+1}$ hence
  $-\lim(s_M) +a_1 = \lim(a_{M+1})$.So $A=-S+a_1$.

I believe this is your attempt at "$\sum (a_i - a_{i+1})$ converges $\implies a_n$ converges". This is mostly fine, but you should be careful when you write $\lim(a_{M+1})$ - you're trying to show that this exists, and here you're assuming it exists. I would be somewhat more careful and write this as
$$a_{M+1} = a_1 - s_M \implies a_M \to a_1 - S$$
