# 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?

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You could let the sequence be your favorite bijection $\mathbb N\to\mathbb Q$. Then every real number is a limit point! –  Henning Makholm Nov 24 '13 at 5:21

You can use something like :{ $1/n$}$\cup${$1+1/n$} $\cup ....\cup$ {$k+1/n$} $\cup....$
What you have written is the set {$k+1/n:k\in\Bbb N$}. How does this define a sequence? –  John Bentin Nov 24 '13 at 14:06
But the countable set {$k+1/n:k\in \Bbb N$} doesn't have any limit points. Did you perhaps mean {$k+1/n:k,n\in \Bbb N$}? –  John Bentin Nov 24 '13 at 22:39
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.