Example for an unbounded sequence with an isolated cluster point with $\|x^{k+1}-x^{k}\|\to 0$

Question

I am looking for an example of an unbounded sequence $(x^{k})_{k\in {\mathbb N}}$ such that $\|x^{k+1}-x^{k}\|\to 0$ and $(x^{k})_{k\in {\mathbb N}}$ has at least one isolated cluster point x but the sequence is not convergent. Can this happen?

Let me clarify what seems to go wrong when trying to prove that $(x^k)_{k\in{\mathbb N}}$ is convergent through contradiction. Lets assume the contrary that $x^k$ is not convergent. Then there exists $\epsilon>0$ and a set of infinite subindices $n_k$ such that $\|x^{n_k} - x\| > \epsilon>0$. Moreover, we know that there exists another set of indices $p_l$ such that $\|x^{p_l}-x\|\to 0$ as $l\to \infty$ (since $x$ is a cluster point). Let $t(n_k)$ denote the smallest index larger than $n_k$ such that $t(n_k)= p_l$ for some $l$. Then using the triangular inequality \begin{equation} \epsilon < \|x^{n_k}-x\| \leq \underbrace{\|x^{t(n_k)}-x\|}_{\to 0} + \underbrace{{\color{red}\sum_{i={n_k}}^{t(n_k)-1}} \underbrace{\|x^{i+1}-x^i\|}_{\to 0}}_{\nrightarrow 0} \end{equation} The issue is the last term where even though individual terms $\|x^{i+1}-x^i\|$ converge to $0$, it is possible that the distant between $t(n_k)$ and $n_k$ migh keep growing as $k\to \infty$ preventing the sum from converging to zero.

Did I miss something? Could you think of an example?

Thank you.

Answer 1

Consider the harmonic series which has step size approaching zero but diverges. We'll make $0$ a cluster point by manipulating this series but also take the time to ensure it remains unbounded. To do this start at some big number with $M_1=1$ then approach $0$ to within some initial tolerance $\epsilon > 0$ by subtracting terms until stepping over $0$, then adding terms to step back, and repeating indefinitely until complete. This is all done after a finite number of terms at which point we simply add the terms until we exceed $M_1+1$ calling that new value $M_2 > M_1$. Now we repeat taking steps back down to zero except we ensure that we come with $\epsilon /2$ of $0$.

Repeating in this fashion we will always have some $M_{k+1}=M_k +1$ larger than some arbitrary $M$ and always have some subsequence that converges to by considering the sequence that arrives within $\epsilon/2^k$ of $0$ while the step size converges to zero.

Answer 2

Having the sequence :

$$ x_k= \sum_{i=1}^k \dfrac{1}{i} e^{j\sum_{i=1}^k 1/i} $$

where $j^2=-1$, verifies the given conditions.

Answer 3

In short, if you have a sequence $(x_k)_{k\geq 0}$ of real numbers such that $|x_{k+1}-x_k| \to 0,$ then the set of cluster points is the interval $[\liminf_k x_k, \limsup_k x_k]$.

If you assume $x_k$ doesn't converge, this means no cluster point can be isolated.

In other words, no such example exists for sequences in $\mathbb{R}$.

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I don't know what may happen when $x_k \in \mathbb{R}^n$ for $n \geq 2$ or $x_k \in \mathbb{R}^\mathbb{R}$.

Answer 4

You can do this with real numbers walking between $0$ and $1$. Let first $x_0=0$, $x_1=1$, $x_2= 1/2,x_3=1/4,x_5=0, x_6,1/8, x_7=1/4, x_8=1/4+1/8, x_9=1/4+2/8, ..$. So you are walking between $0$ and $1$ more and more slowly. You make 2^k step of length $1/2^k$ iup, then down, and when you arrive to $0$ you go to one with speen $1/2^{k+1}$..