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 $\varepsilon &gt; 0$, there are $N_1, N_2$ such that
\begin{align*}
&amp;|a_{2k} -a | &lt; \varepsilon, \text{ if }2k \ge N_1, \\
&amp;|a_{2k+1} -a | &lt; \varepsilon, \text{ if }2k+1 \ge N_2.
\end{align*}
So if $n \ge \max(N_1, N_2)$, we get $|a_n -a| &lt; \varepsilon$.
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 $$a_k = \begin{cases} 0, &amp; \text{ if } n=2^k, k = 0, 1, \dots , \\
1 &amp; \text{ otherwise. } \end{cases}$$ 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 
$$ a_n = \begin{cases} 0 &amp; \text{ if $n$ is prime }, \\ 1 &amp; \text{ if $n$ is not prime.} \end{cases}$$
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.
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 $a_{2k}\to a, \forall\varepsilon&gt;0, \exists n_1\in\Bbb N$ such that $|a_{2k}-a|&lt;\varepsilon, \forall2k\ge n_1$

Since $a_{2k+1}\to b,\exists n_2\in\Bbb N$ such that $|a_{2k+1}-b|&lt;\varepsilon, \forall2k+1\ge n_2$ 

Let us assume $a_n$ converges. This means $a_n$ is Cauchy, or $\forall \varepsilon_1&gt;0,\exists n_3\in\Bbb N$ such that $|x_n-x_m|&lt;\varepsilon_1\forall n,m\ge n_3$. 

Chose $n=m-1=2k\ge\max\{n_1,n_2,n_3\},\varepsilon=\varepsilon_1=\frac14|a-b|&gt;0$

$|a_{2k}-a_{2k+1}|&lt;\frac14|a-b|$ from the convergence of $a_n$, but 

$a_{2k}\in(a-\varepsilon, a+\varepsilon), a_{2k+1}\in(b-\varepsilon,b+\varepsilon)\implies |a_{2k}-a_{2k+1}|&gt;\varepsilon_1$

which is a contradiction.