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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.

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    Start by showing that the limit of the sequences $\{x_{2n}\}$, $\{x_{2n-1}\}$, and $\{x_{3n}\}$ all have the same limit. Either that or (if you're familiar with Cauchy sequences) show that the fact that those three subsequences are Cauchy implies the original sequence is Cauchy, which implies it is convergent. – 2017-02-17

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$\{x_{6n}\}$ 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.

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    Can I just do $n > N_1$, and $n > N_2$ ? instead of $2_n> N_1$ and $2_n+1>N_2$ ? after that $N = max(N_1,N_2)$ – 2017-02-17
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I'll give you the general idea and let you fill in the details.

  1. 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 $L:=L_e=L_o$.
  2. Using the fact that the sequence $(x_{3n})_n$ converges, prove that the even and odd sequences must converge to the same limit.
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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.

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    Can I just do $n > N_1$, and $n > N_2$ ? instead of $2_n> N_1$ and $2_n+1>N_2$ ? after that $N = max(N_1,N_2)$ – 2017-02-17
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    well not really, because we are using the convergence of the sequences $x_{2n}$ and $x_{2n+1}$. – 2017-02-17
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    Can I do like this .. ? Let $\epsilon > 0 $ Since ${x_{2n}}$ is convergent, there exist $N_1$ such that l$x_2n - l $l < $\epsilon$ for all 2n >= $N_1$. Also, Since ${x_{2n-1}}$ is convergent, there exist $N_2$ such that l$x_{2n-1} - l $l < $\epsilon$ for all 2n-1 >= $N_1$. Let N = max{$N_1$,$N_2$} then for n >= N we have l$x_{2n-1} - l $l < $\epsilon$, l$x_2n - l $l < $\epsilon$. Hence l$x_n - l $l < $\epsilon$ for n >= N. – 2017-02-17
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    I didn't do for $x_{3n}$, but I can add it. I just want to know that my way of solving the problme is rright or not. – 2017-02-17
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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.