Example of an $(a_n)$ sequence with exactly $k$ limit points 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.
 A: How about this one?
$$
(a_n)_{n\in N}=n \pmod k
$$
A: 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 $\{a^i\}$.
A: Example:
$$
a_n=\sin\left(\frac{2n+1}{k}\pi\right),\quad n\in\mathbb N.
$$
A: 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 $a_i = a_{f(i)}$
