sequence with infinitely many limit points I am looking for a sequence with infinitely many limit points. I don't want to use $\log,\sin,\cos$ etc.!
It's easy to find a sequence like above, e.g. $1,1,2,1,2,3,1,2,3,4,1,2,3,4,5,1,\dots$
But how can you prove the limit points? The problem I am having is the recursion or definiton of the sequence which I can't name exactly. But for a formal proof I need this.
So what's a sequence with infinitely many limit points without using $\log,\sin,\cos$ or any other special functions?
 A: You can use something like :{ $1/n$}$\cup${$1+1/n$}  $ \cup ....\cup$ {$k+1/n$} $\cup.... $
A: Let $\{r_n\}$ be the set of all rationals. Then sequence $\{r_1, r_1, r_2, r_1, r_2, r_3, r_1, r_2, r_3, r_4,r_1,r_2,r_3,r_4,r_5,...\}$ has all rationals as limit points. 
A: I realize this question was asked a long time ago, but for posterity, you may formally describe the sequence $<1, 1, 2, 1, 2, 3, 1, 2, 3, 4, ...>$ as $<x_i>$ where $x_i = n$ precisely when $i = \frac {(n^2 + (2k-1)n + (k^2 - 3k + 2))}{2}$ for some $k \in \mathbb{N}$. 
To justify this formula, we start by showing that $n$ first appears in the sequence in the term with index $\frac{n(n+1)}{2}$. By splitting the sequence into $(1),(1,2),(1,2,3),(1,2,3,4),...,(1,...,n)$, it is pretty clear that $n$ has index $1+2+3+4+...+n$, which can be shown to equal $\frac{n(n+1)}{2}$ by induction.
Then we continue this useful grouping of the sequence to show the indices of the subsequent terms. $(1),...,(1,...,n),(1,...,n,n+1),(1,...,n,n+1,n+2),...$ The the second appearance comes $n$ terms after the first appearance, the third appearance comes $(n+1)$ terms after the second, and so on. In general the $k^{th}$ appearance of $n$ will come $n + (n+1) + ... + (n+k-2)$ terms after the first appearance. This can be easily shown to be $\frac{(k-2)(k-1)}{2} - \frac{(n-1)n}{2}$. 
But then the $k^{th}$ appearance of $n$ in the sequence will have index $\frac{n(n+1)}{2}+\frac{(k-2)(k-1)}{2} - \frac{(n-1)n}{2}$, which simplifies to $\frac {(n^2 + (2k-1)n + (k^2 - 3k + 2))}{2}$. I'll leave it to you to show that this expression defines the rule of assignment for a bijection  $f:\mathbb{N}\times\mathbb{N} \to \mathbb{N}$, meaning every index $i \in \mathbb{N}$, is mapped to by exactly one pair $(n,k)\in \mathbb{N} \times \mathbb{N}$, therfore showing that $x_i$ is well defined.
Once you have defined this sequence, showing it has infinitely many limit points is easy. We say that $m$ is a limit point of $<x_i>$ precisely if there is a subsequence of $<x_i>$ converging to $m$. Using $f(n,k) = \frac {(n^2 + (2k-1)n + (k^2 - 3k + 2))}{2}$ as our choice function, we choose the subsequence $<y_i>$ where $y_i = x_{f(m,i)} = m$. It stands that $<y_i>$ converges to $m$.
