Krdytkyu

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 $|a_n -a| < varepsilon$.

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.

This is not true. Consider the sequence defined by $a_{2k}=0$ and $a_{2k+1}=1$.

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.

Convergence of Two of the Subsequences

Consider the following three sequences
$$
a_kleft{begin{array}{}0&text{if }kequiv0pmod2\1&text{if }kequiv1pmod2end{array}right.
$$
$limlimits_{ktoinfty}a_{2k}=0$ and $limlimits_{ktoinfty}a_{2k+1}=1$.
$$
b_kleft{begin{array}{}0&text{if }knotequiv1pmod6\1&text{if }kequiv1pmod6end{array}right.
$$
$limlimits_{ktoinfty}b_{2k}=0$ and $limlimits_{ktoinfty}b_{3k}=0$, but $limlimits_{ktoinfty}b_{6k+1}=1$.
$$
c_kleft{begin{array}{}0&text{if }knotequiv2pmod6\1&text{if }kequiv2pmod6end{array}right.
$$
$limlimits_{ktoinfty}c_{2k+1}=0$ and $limlimits_{ktoinfty}c_{3k}=0$, but $limlimits_{ktoinfty}c_{6k+2}=1$.

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.

Let $a_{2k}to a, a_{2k+1}to bne a$. Now since $a_{6k}to a$ and $a_{6k+3}to bne a, a_{3k}$ is not convergent.

Since $a_{2k}to a, forallvarepsilon>0, exists n_1inBbb N$ such that $|a_{2k}-a|<varepsilon, forall2kge n_1$

Since $a_{2k+1}to b,exists n_2inBbb N$ such that $|a_{2k+1}-b|<varepsilon, forall2k+1ge n_2$

Let us assume $a_n$ converges. This means $a_n$ is Cauchy, or $forall varepsilon_1>0,exists n_3inBbb N$ such that $|x_n-x_m|<varepsilon_1forall n,mge n_3$.

Chose $n=m-1=2kgemax{n_1,n_2,n_3},varepsilon=varepsilon_1=frac14|a-b|>0$

$|a_{2k}-a_{2k+1}|<frac14|a-b|$ 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.