TITLE: $(u_{2n})$,$(u_{2n+1})$,$(u_{3n+1})$ converge $\underset{???}{\Rightarrow}(u_n)$ converges
QUESTION [5 upvotes]: Let $(u_n)_{n_\in\mathbb{N}}\in\mathbb{C}^\mathbb{N}$.
We know that $(u_{2n})$, $(u_{2n+1})$ and $(u_{3n+1})$ converge.
The question is to know whether $(u_n)_{n\in\mathbb{N}}$ converges.

/!\ I am not looking for the detailed answer to this question, please do not post it.
Something is bugging me in the above question. If a sequence converges then all its sub-sequences converge towards the same limit.

Therefore we have $l\in\mathbb{C}$ such that :
Let $e>0,\exists n_0\in\mathbb{N},\forall n\geq n_0, |U_{2n}-l|\leq
 e\text{ and }|U_{2n+1}-l|\leq e$
Thus $\forall n\geq 2n_0+1, |U_n-l|\leq e$
Therefore $(u_n)_{n_\in\mathbb{N}}$ converges.

What is bugging me in the above is that we didn't use the convergence of $(u_{3n+1})$ in the proof. Did I miss something ?

REPLY [5 votes]: Yes.  What you missed is that $u_{2n}$ and $u_{2n+1}$ can converge without converging to the same thing.  You need to show that, if they converge to different things, then $u_{3n+1}$ doesn't converge.

REPLY [3 votes]: $(u_{2n})$ and $(u_{2n+1})$ might converge to two different numbers, so that the sequence as a whole does not converge.  But the third sequence alternates between even and odd indices, so it would diverge in such a case.