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?

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    $\begingroup$ In $\mathbb{R}^n$ Cauchy implies convergency because $\mathbb{R}^n$ is a complete metric space. $\endgroup$ May 8, 2017 at 5:04
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    $\begingroup$ In $\mathbb{R}^n$, Cauchy implies convergent, so in order to find a counterexample to Cauchy implies convergent you will have to work in some other space. You don't have to get too exotic; there is a space quite similar (in certain respects) to $\mathbb{R}^n$ that will work. $\endgroup$ May 8, 2017 at 5:16
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    $\begingroup$ In $\mathbb{R}\setminus\{0\}$ consider the sequence $x_n=\displaystyle\frac{1}{n}$. It is Cauchy but not converges. The problem here is because $\mathbb{R}\setminus\{0\}$ is not complete. $\endgroup$ May 8, 2017 at 5:17
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    $\begingroup$ The statement wasn't convergence implies cauchy. It was convergence implies difference of adjancent terms converge. SO the converse counter example is not find a cauchy sequence that doesn't converge (which is impossible) but find a sequence whose difference of adjacent terms converges but the sequence does not. cough cough harmonic cough $\endgroup$
    – fleablood
    May 8, 2017 at 5:51
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    $\begingroup$ Cauchy means for any $e>0$ there is an $M$ so that for ALL $n, m > M$ then $|a_n + a_m| < e$. $a_n$ and the imediately adjacent term $a_{n+1}$ are not ALL $n, m > M$. There is one very famous counter example. $\endgroup$
    – fleablood
    May 8, 2017 at 5:57

1 Answer 1


"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 $n,m > M$.


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