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It is a well-known result that the sequence $$ a_n= \frac{(-1)^nn}{n+1}, $$ has two limit points, and these are $1$ and $-1$.

I'm just looking for some examples of sequences that have exactly $k$ limit points. Of course the sequence will somehow depend on $k$. I'm just curious.

Rustyn
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user153012
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    The answers below are fine but, in case the formulas are distracting, let me just clarify that the underlying idea is just that a periodic sequence such as $5,8,2,5,8,2,5,8,2,\ldots$ has exactly the set ${5,8,2}$ as limit points. It should be clear how to write down a sequence having any given finite set as limit points. – Mike F Sep 30 '14 at 22:50
  • @Yiorgos S. Smyrlis Okay I don't want to rollback, but I don't know why was the original formatting wrong, and why is it neccessary a topic of real analysis. – user153012 Oct 01 '14 at 02:35

4 Answers4

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How about this one? $$ (a_n)_{n\in N}=n \pmod k $$

Ben Grossmann
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Marm
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In general, you can construct a sequence with $k$ limit points by choosing $k$ sequences $(a^i_j)_{j=1}^\infty$, where $i$ indexes which sequence, converging to distinct points $a^i$ (i.e. $a^i_j\underset{j\to\infty}{\longrightarrow} a^i$); then the sequence $(a^1_1,a^2_1,\ldots,a^k_1,a^1_2,\ldots,a^k_2,\ldots)$ will be a sequence with limit points $\{a^i\}$.

Dom
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Example: $$ a_n=\sin\left(\frac{2n+1}{k}\pi\right),\quad n\in\mathbb N. $$

Yiorgos S. Smyrlis
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In this example, $\mathbb{N} = \{1,2,\ldots\}$

The easiest example I could think of:

Take this set:$$\left\{\frac{1}{n} +j: 1\le j \le k, j\in \mathbb{N}, n\in \mathbb{N} \right\}$$ Define $a_{(j,n)} = \frac{1}{n} + j$.

Give the set $\textbf{K} \times \mathbb{N} = \{1,2,3,4,5,\ldots k\}\times \mathbb{N}$ the order: $<$ where $(a,b) < (c,d)$ iff $b<d \text{ or } b=d \text{ and } a<c$. This orders $\textbf{K}\times \mathbb{N}$ in type $\mathbb{N}$ so that we can define a function $f:\mathbb{N}\to \textbf{K} \times \mathbb{N}$ $$ f(0) = {\min\left(\textbf{K} \times \mathbb{N} , <\right)}\\ f(i+1) = \min\left(\textbf{K}\times \mathbb{N}\setminus f(i), <\right) $$ Now set $a_i = a_{f(i)}$

Rustyn
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  • To nitpick: this isn't a sequence unless you specify some ordering of the terms (since your terms are really of the form $a_{jn}$ with two indices rather than simply $a_n$). Any ordering does work, but they still do need to be ordered. – Steven Stadnicki Sep 30 '14 at 23:00
  • @StevenStadnicki I fixed my answer, are you OK with it now? – Rustyn Oct 01 '14 at 08:57
  • I think it's fine, although the ordering could probably be described more explicitly - it's a little clunkier than the other two but has the advantage of having (given the right parameters) no repeated terms. – Steven Stadnicki Oct 01 '14 at 15:27
  • @StevenStadnicki How about now? You're right, an explicit ordering is easy to write down in this case, and preferable. The ordering described above yields: $(1,1)<(2,1)<\ldots <(k,1)<(1,2)<(2,2)<\ldots <(k,2)<\ldots$ which is what we want, because now we've an ordering of $\textbf{K}\times \mathbb{N}$ in type $\mathbb{N}$. – Rustyn Oct 01 '14 at 16:54