Consider the sequence $\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},\ldots$ Consider the sequence 
$$\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},\ldots$$ For which numbers $b$ is there a subsequence converging to $b$?
 A: Your sequence includes every rational number in $(0,1)$ infinitely many times. For any $b\in[0,1]$, there exists a sequence of rational numbers in $(0,1)$ converging to $b$. That sequence necessarily occurs as a subsequence of your sequence. Conversely, any $b\notin[0,1]$ cannot have a subsequence of your sequence converge to it, as there exists a positive number $\epsilon$ such that every element of your sequence is more than $\epsilon$ away from $b$.
Thus, the numbers $b$ for which there is a subsequence converging to $b$ are precisely those $b\in[0,1]$.
A: Anon pointed out $[0,1]$ as the answer : your sequence goes like this :
$$
\dots, \frac {n-2}{n-1}, \frac 1n, \frac 2n, \dots, \frac {n-1}n, \frac 1{n+1}, \dots
$$
Now every element in $[0,1]$ is in one of the intervals $[0,1/n), [1/n,2/n), \dots, [n-1/n,1]$. Let $\alpha \in [0,1]$ and define $x_n$ to be the lower bound of the interval in which $\alpha$ lies in. Since the length of those intervals is $1/n$, $|x_n - \alpha| \le 1/n \to 0$ as $n \to \infty$ and $x_n$ is a well-defined subsequence of your sequence (it preserves the ordering and everything). 
You can't get any number outside $[0,1]$, since they are not limit points of the set of points in your sequence. Therefore the answer you're looking for is $[0,1]$.
Hope that helps,
