Convergence of a sequence in R If $a_n$ is a sequence of real numbers, then I know the following result: 
If    $\underset{n\to\infty}{\lim} a_{n}$ exist then $$\lim_{n\rightarrow\infty} a_{2n} \quad \text{and} \quad \lim_{n\to\infty} a_{2n-1} \quad \text{and} \quad  \lim_{n\rightarrow\infty} a_{3n}$$ exist and equal. Now what can I say about the converse? i.e.
if I know that $$\lim_{n\rightarrow\infty} a_{2n} \quad \text{and} \quad \lim_{n\to\infty} a_{2n-1} \quad \text{and} \quad  \lim_{n\rightarrow\infty} a_{3n}$$ exist, what can i say about $\underset{n\to\infty}{\lim} a_{n}$?
 A: $(a_{3n})$ contains a subsequence , which is also a subsequence of $(a_{2n})$ 
(which ?). 
Hence $(a_{3n})$ and $(a_{2n})$ has the same limit.
$(a_{3n})$ contains another subsequence , which is also a subsequence of $(a_{2n-1})$ 
(which ?). 
Hence $(a_{3n})$ and $(a_{2n-1})$ has the same limit.
(which ?). 
Your turn !
A: Let $\alpha := \lim_n a_{2n}$, $\beta := \lim_n a_{2n-1}$ and $\gamma := \lim_n a_{3n}$. Note that $(a_{6n})$ is a subsequence of $(a_{2n})$, hence $a_{6n} \to \alpha$, but $(a_{6n})$ is also a subsequence of $(a_{3n})$, therefore $a_{6n} \to \gamma$, hence $\alpha = \gamma$. Along the same line of thought, considering $(a_{6n+3})$, we see that $\gamma = \beta$, therefore $\alpha = \beta$.
Hence $(a_{2n})$ and $(a_{2n-1})$ both converge to $\alpha$, starting from this we will prove $a_n \to \alpha$. Let $\epsilon > 0$. As $a_{2n}\to \alpha$, there is $N_1$ such that 
$$ |a_{2n} - \alpha|< \epsilon, \quad n \ge N_1 $$
Analogously, for some $N_2$, we have 
$$ |a_{2n-1} - \alpha|< \epsilon, \qquad n \ge N_2 $$
Now let $N := \max\{2N_1, 2N_2 - 1\}$ and $n \ge N$. If $n$ is even, $n = 2k$ with $k \ge N_1$, if $n$ is odd, $n = 2k-1$ with $k \ge N_2$. In both cases 
$$ |a_n - \alpha|< \epsilon. $$
As this holds for all $n\ge N$, $a_n \to \alpha$.
