The growth of a sequence related to Liouville numbers [closed]

Question

I am doing a work on Liouville numbers. The Liouville constant $\ell=\sum_{k\geq 0}10^{-k!}$ has its approximation by rational numbers related to the fact that for $v_n=n!$, then $v_{n+1}/v_n$ tends to infinity as $n\to \infty$.

I am try to construct a more special class of Liouville numbers and for that I would like to construct an integer sequence $v_n$ such that

$c_1n<v_{n+1}/v_n<c_2n$ and $d_1n<v_{2n+2}/v_{2n}<d_2n$ (for some positive constants $c_1,c_2, d_1, d_2$).

See that for $v_n=n!$ this does not happen, since $v_{n+1}/v_n=n+1$ while $v_{2n+2}/v_{2n}=O(n^2)$.

Someone please may help me in this task?

Answer 1

Not sure whether I understand correctly, but can you just take: $$v_n = \sqrt{n!}$$ (or the integer part of it)?

You will then have $v_{n+1}/v_{n}=O(\sqrt{n})$, which is still $O(n)$.

EDIT:

Now that the question has been edited, it really requires $\Theta(n)$ instead of $O(n)$.

But then the answer is simply impossible:

If we have $a n\leq v_{n+1}/v_{n}$ for some constant $a$ and all sufficiently large $n$, then it follows that $$v_{2n+2}/v_{2n}= v_{2n+2}/v_{2n+1} \times v_{2n+1}/v_{2n}\geq a^2 (2n)(2n+1) \geq 4a^2 n^2.$$ Hence it cannot be $O(n)$.