A problem on convergent subsequences Find a sequence $\langle X_n\rangle$ such that
$$L_{X_n}=\left\{\frac{n+1}{n}:n\in\mathbb N\right\}\cup \{1\}.$$
Where $L_{X_n} = \{ p\in \mathbb{R} : \text{There exists a subsequence } \langle X_{n_k}\rangle \text{ of } \langle X_n\rangle \text{ such that} \lim X_{n_k} = p\}$.
Justify your answer.
Since given $L_{X_n}$ is an infinite set, it was difficult for me to find a sequence in which I could select infinite ordered unending set of real numbers ( Subsequences) in such a way that they converge to the given real numbers in $L_{X_n}$.
Can anyone please help me in order to solve this? 
*An edit was made to define $L_{X_n}$
 A: Pick a set of sequences $a_n(m)$ that have limits $(n+1)/n$, or $a_0$ has limit $1$.  Then interleave them.  Remember the proof that the rationals are countable.
A: For example $$\left( 2, 2,\frac{3}{2},  2,\frac{3}{2} , \frac{4}{3},2,\frac{3}{2} , \frac{4}{3} ,\frac{5}{4} ,...\right)$$
A: The easiest solution to this problem is to construct a sequence such that each element in your set $L_{X_n}$ (aside from 1) appears infinitely many times. That way there is a constant subsequence, which will obviously converge to the thing that you want it to.
Here is a sequence that ought to have the needed property. I'm hiding it with a spoilers tag in case you want to try and work it out on your own.

 $\left\{ \frac{2}{1},\frac{2}{1},\frac{3}{2},\frac{2}{1},\frac{3}{2},\frac{4}{3},\frac{2}{1},\frac{3}{2},\frac{4}{3},\frac{5}{4},\frac{2}{1},\frac{3}{2},\frac{4}{3},\frac{5}{4},\frac{6}{5},\frac{2}{1},\frac{3}{2},\frac{4}{3},\frac{5}{4},\frac{6}{5},\frac{7}{6},\frac{2}{1},\frac{3}{2},\frac{4}{3},\frac{5}{4},\frac{6}{5},\frac{7}{6},\frac{8}{7} \dots \right\}$

