Sequence $x_n$ has subsequences $\{x_{2n}\}$, $\{x_{2n-1}\}$, $\{x_{3n}\}$ all converging. show that $\{x_{n}\}$ is convergent. Suppose that $x_n$ is a sequence such that the subsequences $\{x_{2n}\}$, $\{x_{2n-1}\}$, $\{x_{3n}\}$ all converge. Show that $\{x_{n}\}$ is convergent. 
If there is an actual sequence of $\{x_n\}$, I might be able to solve it. But I don't know how to start and solve this problem. 
 A: $\{x_{6n}\}$ is a subsequence of $\{x_{2n}\}$ and $\{x_{3n}\}$ so it converges to the same limit as both of them.  Similarly, $\{x_{6n+3}\}$ is a subsequence of the odds and $\{x_{3n}\}$ so it converges to the same limit as both of them.  So all three sequences converge to the same limit.
For every $\epsilon>0$ we can find $N_1$ and $N_2$ such that if $2n>N_1$ then $|x_{2n}-l|<\epsilon$ and if $2n+1>N_2$ then $|x_{2n}-l|<\epsilon$;thus taking $N=\max(N_1,N_2)$ works.
A: I'll give you the general idea and let you fill in the details.


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*By assumption, the sequences of even and odd terms converge to some limits $L_e$ and $L_o$. Show that, if $L_e = L_o$, then the whole sequence $(x_n)_n$ converges to $L:=L_e=L_o$.

*Using the fact that the sequence $(x_{3n})_n$ converges, prove that the even and odd sequences must converge to the same limit.

A: If we had that $x_{2n}$ and $x_{2n+1}$ converged to the same value $l$ it would be easy. Because for every $\epsilon>0$ we could find $N_1$ and $N_2$ such that if $2n>N_1$ then $|x_{2n}-l|<\epsilon$ and if $2n+1>N_2$ then $|x_{2n}-l|<\epsilon$, taking $N=\max(N_1,N_2)$ would do the trick.
So now we just have to show that $x_{2n}$ and $x_{2n+1}$ converge to the same value. To do this pick a common subsequence of $x_{3n}$ and $x_{2n}$ and notice it converges to the limit of $x_{2n}$ and to the limit of $x_{3n}$. So these two must be equal. Then pick a common subsequence of $x_{2n+1}$ and $x_{3n}$ and do the same thing.
A: Hint: Note that $x_{3n}$ has two subsequences such that one is also a subsequence of $x_{2n}$ and the other a subsequence of $x_{2n-1}$. And we know that  a converging sequence has the same limit as every subsequence. And from there it is straight forward. 
