Any two of the subsequences $\{a_{2k}\}, \{a_{2k+1}\}, \{a_{3k}\}$ converge do not imply $\{a_k\}$ converges? If $\{a_k\}$ is a sequence, I have shown if the three subsequences $\{a_{2k}\}, \{a_{2k+1}\}, \{a_{3k}\}$ all converge, then the sequence converges.
I am asked whether the sequence converges if only two of them converge. Any help will be appreciated. 

I first show the limits are the same. Let us assume $\lim a_{2k} = a, \lim a_{2k+1}=b, \lim a_{3k}=c$. Then $\lim_{6k} = a = c$ and $\lim_{6k +3} = b = c$. Now pick any $\varepsilon > 0$, there are $N_1, N_2$ such that
\begin{align*}
&|a_{2k} -a | < \varepsilon, \text{ if }2k \ge N_1, \\
&|a_{2k+1} -a | < \varepsilon, \text{ if }2k+1 \ge N_2.
\end{align*}
So if $n \ge \max(N_1, N_2)$, we get $|a_n -a| < \varepsilon$.
 A: The answer is no. I remember I learned the example from my analysis class.
Take $a_n = (-1)^n$, then $\lim_{k \to \infty} a_{2k} =1$ and $\lim_{k \to \infty} a_{2k-1} = -1$.
If $$a_k = \begin{cases} 0, & \text{ if } n=2^k, k = 0, 1, \dots , \\
1 & \text{ otherwise. } \end{cases}$$ Then $\lim_{k \to \infty} a_{3k}  = 1 $ and $\lim_{k \to \infty} a_{2k+1} = 1$ but $\lim_{k \to \infty} a_{2k}$ does not exist.
Finally, if 
$$ a_n = \begin{cases} 0 & \text{ if $n$ is prime }, \\ 1 & \text{ if $n$ is not prime.} \end{cases}$$
We can check $\lim_{k \to \infty} a_{3k} = 1$ and $\lim_{k \to \infty} a_{2k} =1 $ but $\lim_{k \to \infty} a_{2k+1}$ does not exist.
A: This is not true. Consider the sequence defined by $a_{2k}=0$ and $a_{2k+1}=1$. 
A: Let $(a_n)$ defined by
$$a_{6k}=1$$
$$a_{6k+1}=a_{6k+5}=0$$
$$a_{6k+2}=1$$
$$a_{6k+3}=1$$
$$a_{6k+4}=1$$
then
$(a_{3n})$ and $(a_{2n})$ converge but
$(a_n)$ does not.
A: Convergence of Two of the Subsequences
Consider the following three sequences
$$
a_k\left\{\begin{array}{}0&\text{if }k\equiv0\pmod2\\1&\text{if }k\equiv1\pmod2\end{array}\right.
$$
$\lim\limits_{k\to\infty}a_{2k}=0$ and $\lim\limits_{k\to\infty}a_{2k+1}=1$.
$$
b_k\left\{\begin{array}{}0&\text{if }k\not\equiv1\pmod6\\1&\text{if }k\equiv1\pmod6\end{array}\right.
$$
$\lim\limits_{k\to\infty}b_{2k}=0$ and $\lim\limits_{k\to\infty}b_{3k}=0$, but $\lim\limits_{k\to\infty}b_{6k+1}=1$.
$$
c_k\left\{\begin{array}{}0&\text{if }k\not\equiv2\pmod6\\1&\text{if }k\equiv2\pmod6\end{array}\right.
$$
$\lim\limits_{k\to\infty}c_{2k+1}=0$ and $\lim\limits_{k\to\infty}c_{3k}=0$, but $\lim\limits_{k\to\infty}c_{6k+2}=1$.
For a sequence to converge, all subsequences must converge to the same limit.
Thus, the convergence of two of the subsequences does not imply the convergence of the whole sequence.
A: Let $a_{2k}\to a, a_{2k+1}\to b\ne a$. Now since $a_{6k}\to a$ and $a_{6k+3}\to b\ne a, a_{3k}$ is not convergent.
Since $a_{2k}\to a, \forall\varepsilon>0, \exists n_1\in\Bbb N$ such that $|a_{2k}-a|<\varepsilon, \forall2k\ge n_1$
Since $a_{2k+1}\to b,\exists n_2\in\Bbb N$ such that $|a_{2k+1}-b|<\varepsilon, \forall2k+1\ge n_2$ 
Let us assume $a_n$ converges. This means $a_n$ is Cauchy, or $\forall \varepsilon_1>0,\exists n_3\in\Bbb N$ such that $|x_n-x_m|<\varepsilon_1\forall n,m\ge n_3$. 
Chose $n=m-1=2k\ge\max\{n_1,n_2,n_3\},\varepsilon=\varepsilon_1=\frac14|a-b|>0$
$|a_{2k}-a_{2k+1}|<\frac14|a-b|$ from the convergence of $a_n$, but 
$a_{2k}\in(a-\varepsilon, a+\varepsilon), a_{2k+1}\in(b-\varepsilon,b+\varepsilon)\implies |a_{2k}-a_{2k+1}|>\varepsilon_1$
which is a contradiction.
