# Let $(a_n)$ be a sequence of real numbers. Suppose each of the subsequences $(a_{2n}),(a_{2n+1})$ and $(a_{3n})$ converges to $p,q$ and $r$

Let $(a_n)$ be a sequence of real numbers. Suppose each of the subsequences $(a_{2n}),(a_{2n+1})$ and $(a_{3n})$ converges to $p,q$ and $r$, respectively. Show that $p=r$ and $q=r$ and hence conclude that $(a_n)$ converges. To prove $p=r$, I consider a subsequence of $(a_{2n})$ and $(a_{3n})$ , which is $(a_{6n})$. Then I say clearly $(a_{6n})$ converges to $p$ and $r$ and hence $p=r$. Is this proof work?

-
Yes, this works perfectly. –  1015 Feb 9 '13 at 14:52
As Julien says, yes; and you can use a similar idea for the second part (cross multiply the indices is probably the easiest way). –  gnometorule Feb 9 '13 at 14:56
And for the other equality, consider for instance $(a_{3(2n+1)}=a_{2(3n+1)+1})$. –  1015 Feb 9 '13 at 14:56

Yes, $a_{6n}$ is a subsequence of $a_{2n}$ and $a_{3n}$, so it converges to both $p$ and $r$, thus $p=r$.
Similarly, $a_{6n+3}=a_{3(2n+1)}=a_{2(3n+1)+1}$ is a subsequence of $a_{3n}$ and $a_{2n+1}$, so it converges to both $r$ and $q$, thus $r=q$.
So $p=r=q$ and, as $a_{2n}$ and $a_{2n+1}$ covers all of $a_n$, it converges.