Is being a Cauchy sequence equivalent to $ \lim_{n\to+\infty}d(x_{n+k},x_n)=0$ for every $k$? Is this statement true? 

In a metric spase $(E,d)$, a sequence $(x_n)$ is Cauchy if and only if $  \forall k\in \mathbb{N}, \lim_{n\rightarrow+\infty}d(x_{n+k},x_n)=0$

I proved that $\Rightarrow$ is true,indeed if $(x_n)$ is Cauchy then for all $\varepsilon>0,\exists n_0\in\mathbb{N}, \forall p,q\geq n, d(x_p,x_q)<\varepsilon$ it still right for all $ n\geq n_0$ and $m=n+k>n\geq n_0$ 
But I don't know if $\Leftarrow$ is true ? 
 A: The equivalence is not correct.
To see this, consider any divergent series of real numbers $\sum a_n$ such that 
$a_n\to 0$ as $n\to \infty$; for example $a_n=\frac1n$. Then define $x_n=a_1+\dots +a_n$. This is a non-Cauchy sequence (as any non-convergent sequence of real numbers), but $x_{n+k}-x_n=a_{n+1}+\dots +a_{n+k}\to 0$ as $n\to \infty$ for any fixed $k\in\mathbb N$, since this is the sum of $k$ terms tending to $0$; that is, $d(x_n,x_{n+k})\to 0$ as $n\to\infty$. So the implication $\Leftarrow$ does not hold.
If you prefer, you can take $x_n=\log n$ or $x_n=\sqrt{n}$, and check that $x_{n+k}-x_n\to 0$ as $n\to \infty$, for any $k\in\mathbb N$.
On the other hand, your proof of the other implication is correct.
Last remark : in fact the assumption "$\forall k\in\mathbb N\; d(x_{n+k},x_n)\to 0$" is equivalent to the seemingly weaker "$d(x_{n+1},x_n)\to 0$", as you can easily check.
A: of course that it's wrong !!
Consider $x_n=\frac{1}{n}$, the sequence converge therefore it's a Cauchy sequence, but $d(x_{n+k},x_n)\neq 0$ for all $k$.
The correct statement  would be
$(x_n)$ is cauchy $\iff \forall \varepsilon>0, \exists N\in\mathbb N:\forall n,k\in\mathbb N, n\geq N\implies d(x_{n+k},x_n)<\varepsilon$
and this is the definition of a Cauchy sequence as well !
A: With the way you edited it, you actually just wrote the definition of a cauchy sequence in a different manner.
$\lim_{n\to+\infty}d(x_{n+k},x_n)=0$ means that for every $\varepsilon>0$ there exists $N$ such that $d(x_{n+k},x_n)<\varepsilon\ \forall\ n>N$.
But since $k$ can be any arbitrary natural number, this means that $d(x_n,x_m)<\varepsilon\ \forall\ n>N,m>N$.
Thus, you didn't really change anything there.
