Prove that it is not guaranteed that if some pair of subsequences $\{a_{3n}\}, \{a_{2n}\}, \{a_{2n + 1}\} $ of sequence $\{a_n\}$ are converging... So, we need to prove that if some pair of these subsequences are converging it doesn’t mean that $a_n$ is converging too.
For pair $\{a_{2n}\}, \{a_{2n + 1}\}$ everything is pretty much obvious. So, $a_n = (-1)^n$ would be perfect counterexample.
But how to prove it for two other possible combinations of subsequences? I guess that we should consider remainder of $n$ from division by 6 and see that any of two left combinations can’t cover all remainder. So, let’s create a counterexample considering that.
But, unfortunately, I can’t think of any counterexample. Maybe I was going in wrong direction? Thank you for your answers and hints in advance!
 A: Consider the sequence
$$a_k=\begin{cases}1&k\text{ is prime}\\0&\text{else}\end{cases}$$
Then $\{a_{2n}\}$ and $\{a_{3n}\}$ clearly converge to $0$, but $\{a_n\}$ does not converge because there are infinitely many primes.
For $\{a_{2n+1}\}$ and $\{a_{3n}\}$ replace $k$ with $k+3$ in the formula above.
A: You may take $\color{red}{a_{x_n}}$ and  $\color{blue}{a_{y_n}}$ as follows:
$$a_n = \{\color{red}{-1},\color{blue}1, \color{blue}{-1} \mid \color{red}1,\color{blue}{-1}, \color{red}1,\color{blue}{-1},\color{red}1,\color{blue}{-1}, \color{red}{1} ... \}$$
where after the $\mid$ sign, they are just your counter-example, but we just changed the initial terms.
Of course, we have that $$\lim_{n\to \infty} \color{red}{a_{x_n}} = \color{red}1$$ and $$\lim_{n\to \infty} \color{blue}{a_{y_n}} = \color{blue}{-1}$$
How does it work?
Note that, $x_n$ and $y_n$ can implicitly given as:
$$x_n = 1, 4, 6, 8, 10,... \\
y_n = 2, 3, 5, 7, 9,...
$$
or more precisely as:
$$
x_n = \begin{cases}
1, \ \  n = 1 \\
2n, \ \ n > 1
\end{cases} \ \ \text{ and } \ \ 
y_n = \begin{cases}
2, \ \  n = 1 \\
2n - 1, \ \ n > 1
\end{cases}
$$
A: Let $a_n=(-1)^{m}$ if $n=5^{m}$ and $a_n=0$ for any $n$ which is not of the form $5^{n}$. Then $a_{3n}$ and $a_{2n}$ are convergent but $a_n$ is not.
Let $a_n=(-1)^{m}$ if $n =2^{m}$ and $0$ for other $n$. The $a_{3n}$ and $a_{2n+1}$ are convergent but $a_n$ is not.
