$y_{2n}, y_{2n+1}$ and $y_{3n}$ all converge. What can we say about the sequence $ y_n$? My friend and I are currently debating the following question: 

Let $y_n$ be a sequence in a metric space and assume that the subsequences 
  $y_{2n}$, $y_{2n + 1}$, and $y_{3n}$ all converge. What can we say about the sequence $y_n$? 

To me it seems that all the subsequences will converge to a limit say $y$. Then any term in the original sequence will be in one of these subsequences, it follows that the entire sequence $y_n$ converges to $y$.
I let $\varepsilon > 0$. Since $y_{2n}$ converges to $y$ there exists an index $2N$ such that $d(y_{2n},y) < \varepsilon$ for $2n \geq 2N$,
and since $(y_{2n+1})$ converges to $y$ there exists $2M + 1$ such that $d(y_{2n+1}, y) < \varepsilon$ for $2n + 1 \geq 2M + 1$.
Let $m \geq \max\{2N,2M+1\}$. If  $m$ is even, since $m\geq 2N$ we have $d(y_m,y)<\varepsilon$; if $m$ is odd, since $m\geq 2M+1$ we have $d(y_m,y)<\varepsilon$. Thus $d(y_m,y)<\varepsilon$ for any $m \geq \max\{2N,2M+1\}$, and we conclude that $(y_n)$ converges to $y$. 
However, my friend does not think that this is completely correct because I didn't use the $y_{3n}$ in the proof. I think he is right, that I do in fact need it, but I am unsure of how to add it in. All in all, I think that since any term in the original sequence will be in one of these subsequences, it follows that the entire sequence $(y_n)$ converges to $y$.
$2n+1$ covers all the odd terms and $2n$ converse all the even terms, so is $3n$ even needed? I would like to use it since the proof has it in the question, but I am unsure of what to do. Could someone kindly help me construct this proof in a better manner,and is my line of thinking correct? Thank you! 
 A: You need to use the fact that if a sequence converges, any subsequence converges to the same limit.  Since $\{y_{3n}\}$ converges to say $y$, then $\{y_{6n}\}$ must also converge to $y$.  But $\{y_{6n}\}$ is a subsequence of $\{y_{2n}\}$ and since that sequence converges, it must converges to $y$.  A similar argument will show that $\{y_{2n+1}\}$ will converge to $y$.
A: Let $L$ be the limit of $y_{2n+1}$, $M$ be the limit of $y_{2n}$, and let $N$ be the limit of $y_{3n}$.
Then $y_{6n + 3}$ is a subsequence of both $y_{2n + 1}$ and $y_{3n}$. Thus it converges and the limit of $y_{6n + 3}$ is the same as the limit of $y_{2n + 1}$ and the limit of $y_{3n}$. Hence $L = N$.
Similarly, $y_{6n}$ is a subsequence of both $y_{2n}$ and $y_{3n}$. Thus it converges and the limit of $y_{6n}$ is the same as the limit of $y_{2n}$ and the limit of $y_{3n}$. Hence $M = N$.
So you have $L = M = N$. Now that you know $L = M$, the argument you gave shows that the sequence converges. If you didn't know $L = M$, then what you call $y$ in the argument would not be well defined; it would be one value for the odd sequence and another value for the even sequence.
A: We have $y_{2n}\to L, y_{3n}\to M.$ Now $y_{6n}$ is a subsequence of both of these sequences. Therefore $L=M.$ We also know $y_{2n+1}$ converges to some $N.$ We must have $N=M$ because infinitely many of $\{3n\}$ are odd. So now all of these sequences converge to the same limit. In particular $y_{2n},y_{2n+1}$ converge to the same limit, and that impies $y_n$ converges
