Does the convergence of some subsequences imply the convergence of a sequence? I am considering a following problem: Does 
$$\\a_{2k}, a_{2k+1}, a_{3k} \rightarrow g $$
imply that $$a_{n} \rightarrow g ?$$
I know that if every subsequence goes to $g$ then also a sequence goes to $g$. My way of reasoning is following: we know that odd and even $k$ subsequences goes to $g$. But if we substitute for example $k=t^{2} -5$, then there is problem (at least for me). If I can reason that way, what must I do to prove more formally that the statement is false? And if I can not, what is going on with $a_{n}$? Thanks for any hints!
 A: If the even numbered terms and odd numbered terms both converge to the same limit $L$, then the limit of the sequence exists and is $L$.
To prove this, let $\epsilon > 0$ be arbitrary. We want to find $N \in \mathbb{N}$ such that $n > N$ implies $|a_n - L| < \epsilon$.
Since the even numbered terms are all within epsilon of $L$ after some $N_1 = 2j$ and the odd numbered terms are all within epsilon of $L$ after some $N_2 = 2k + 1$, just pick $N = \max\{N_1, N_2\}$.
A: You may want to prove the next nice lemma:
Lemma: If $\,\{A_i\}_{i\in I},$ is some partition of $\,\Bbb N\,$ s.t. $\,|A_i|=\aleph_0=|\Bbb N|\,\,\,,\,\forall\,\,i\in I\,$ , and all the sets $\,A_i\,$ are well ordered, then for a real sequence $\,\{x_n\}\,$ it is true that
$$x_n\xrightarrow [n\to\infty]{} x\Longleftrightarrow (\,\forall\,\,i\in I\,\,,x_{n_i}\xrightarrow [n_i\to\infty\,,\,n_i\in A_i]{} x)$$
Thus, as B. wrote in his answer, it is enough and sufficient that the subsequence of odd indexes and the one of even indexes converge both to the same limit.
A: In the other answers you have seen that it $a_{2k},a_{2k+1}\to g$ then already $a_n\to g$. You don't even need $a_{3k}$. I would like to add that if you just require $a_{2k},a_{2k+1}$ and $a_{3k}$ to converge (a priori with different limits), then you have that all three limits coincide and that $a_n$ has the same limit.
To see this note that $a_{2n}$ and $a_{3n}$ as well as $a_{2n+1}$ and $a_{3n}$ have a commen subsequence. Then you only need that if a converging sequence has a subsequence with limit $g$ then the whole sequence has limit $g$.
