Proving convergence by convergence of 3 subsequences I'm studying for my real analysis exam and came across this prompt:
Let $\{a_n\}$ be a sequence of real numbers. If the subsequences $\{a_{2k}\}$,
$\{a_{2k+1}\}$and $\{a_{3k}\}$ converge, prove that $\{a_n\}$  converges.
I thought maybe that I could prove $\{a_n\}$ converges by first showing it's a Cauchy sequence, but I have not made much progress. 
I'd really appreciate it if someone could show me an easy way to do this proof or even just give an outline.
 A: HINT: what can you say about $(a_{2k})$ and $(a_{2k+1})$ limits if $(a_{3k})$ converges? 
Let $\lim (a_{2k})=L$,  $\lim (a_{2k+1})=L'$,  $\lim (a_{3k})=L''$. 
Extract $(a_{6k})$,  from $(a_{2k})$ with $k\mapsto 3k$. You then have $\lim (a_{6k})=\lim (a_{2k})=L$. Using $k \mapsto 2k$, $\lim (a_{6k})= \lim (a_{3k})=L''$. As limit is unique, $L=L''$. 
Then extract $(a_{6k+3})$,  from $(a_{2k+1})$ with $k\mapsto 3k+1$. You then have $\lim (a_{6k+3})=\lim (a_{2k+1})=L'$. Using $k \mapsto 2k+1$, $\lim (a_{6k+3})= \lim (a_{3k})=L''$. As limit is unique, $L'=L''$. 
Hence :
$$\lim (a_{3k})= \lim (a_{2k})= \lim (a_{2k+1})=L.$$
Now use the definition of $\lim(u_n)=L$ for $(a_{2n})$ and $(a_{2n+1})$. 
A: Here are some hints:


*

*Lets say that $\lim_{k \to \infty} a_{2k} = a_{even} \in \mathbb{R}$ and $\lim_{k \to \infty} a_{2k+1} = a_{odd} \in \mathbb{R}$. Under what condition does $\lim_{n \to \infty} a_n$ exist?

*What can you say about the subsequences $(a_{6k})$ and $(a_{6k + 3})$ of $(a_{3k})$?
If you really want the complete proof, just leave a comment :)
