Sequence in $\mathbb{R}^n$ convergent implies Cauchy, but not vice versa? Earlier I proved that a sequence in $\mathbb{R}^n$ converges to some limit $\vec{a}$ if and only if it is Cauchy. Now, I've also proved that if the sequence $\{\vec{x}_k\}$ is convergent to some limit $\vec{a} \in \mathbb{R}^n$, then $\lim\limits_{k\to\infty}\|\vec{x}_k - \vec{x}_{k+1} \| = 0$. I proved it in the following way: since $\{\vec{x}_k\}$ is convergent, it is Cauchy, which implies that, $\forall\varepsilon >0, \exists K\in \mathbb{N}$ such that $\|\vec{x}_k-\vec{x}_l\|<\varepsilon$ whenever $k,l\ge K$. Let $k>K,l=k+1$, then $k+1>K$ and $\|\vec{x}_k-\vec{x}_{k+1}\|<\varepsilon$, which implies that $\lim\limits_{k\to\infty}\|\vec{x}_k - \vec{x}_{k+1} \| = 0$ by definition of the limit at infinity.
Now, the next question I have to solve seems to contradict the above. Namely, the question asks me to provide a counterexample in $\mathbb{R}^n$ (for any $n\in\mathbb{N}$) to show that the converse of the statement above is not true. Because how can it be not true if Cauchy implies convergence?
 A: "Because how can it be not true if Cauchy implies convergence?"
Because your counter-example will not/can not be Cauchy.
$\{a_n\}$ being Cauchy/convergent implies $|a_n - a_{n+1}|$ converges to $0$ but $|a_n - a_{n+1}|$ converging to $0$ does  not imply $\{a_n\}$ is Cauchy/convergent.
And that is what you must find a counter example of: $\{a_n\}$ being such that $|a_n - a_{n+1}|$ converges to $0$ but $\{a_n\}$ is not Cauchy/convergent.
======= counter example below ======
Let $a_n = \sum_{i=1}^n \frac 1i$. Then $a_n$ does not converge but $|a_n - a_{n+1}| = \frac 1{n+1} \rightarrow 0$.
Cauchy is all $|a_n - a_m| \rightarrow 0$ for $n,m > M$ for some $M$.  If only $|a_n - a_{n+1}|\rightarrow 0$ that is only the difference of adjacent terms.  It does not imply Cauchy.
Obviously if all $n,m > M$ are such that $|a_n - a_m| < \epsilon$ then for all $n > M$ then $|a_n-a_m| < \epsilon$ and $n+1 > M$.  But the converse isn't necessarily true.  That something is true for all $n, n+1 > M$ in no way should it be true for all $n,m > M$.
