<s>[INST] Formal proof of the sequence involving double factorials $x_n = \frac{(2n)!!}{(2n-1)!!}$ is not bounded. I'm trying to prove the following:

Let $n\in \mathbb N$ and 
  $$ 
x_n = \frac{(2n)!!}{(2n-1)!!}
$$ 
  Where:
  $$
(2n)!! = 2\cdot 4 \cdot 6 \cdot \dots \cdot 2n \\
(2n - 1)!! = 1\cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)
$$
  Prove $\{x_n\}$ is not bounded.

Intuitively it feels like it is unbounded. Let's consider the following fraction:
$$
x_n = \frac{2\cdot 4 \cdot 6 \cdot \dots \cdot (2n)}{1\cdot 3 \cdot 5 \cdot \dots \cdot (2n-1) }
$$
Now if we consequently take a number from nominator and denominator we get:
$$
\frac{2}{1} > 1 \\
\frac{4}{3} > 1 \\
\frac{6}{5} > 1 \\
\dots \\
\frac{2n}{2n-1} > 1
$$
So the fraction is a product of rational numbers each of which is greater than $1$ and the product of rational numbers greater than $1$ is increasing. 
I've tried to formalize that by expanding $(2n)!!$ and $(2n-1)!!$:
$$
(2n)!! = (2n)(2n-2)(2n-4)\cdots(4)(2) = 2^kn!\\
(2n-1)!! = (2n-1)(2n-3)\cdots(5)(3)(1) = \\
= \frac{(2n-1)(2n-2)(2n-3)\cdots(5)(4)(3)(2)(1)}{(2n-2)(2n-4)\dots(4)(2)} = \\
\frac{(2n-1)!}{2^{n-1}(n-1)!}
$$
So using the above:
$$
x_n = \frac{2^nn!2^{n-1}(n-1)!}{(2n-1)!} = \frac{2^{2n-1}n!(n-1!)}{(2n-1)!}
$$
Here is where I got stuck. How do i proceed with the proof using some constant $M$ and some number $N > n$ such that $x_N > M$? Should i introduce some inequality?
I've seen a similar question, but the sequence there is proven to be divergent which i guess is more a calculus concept (yet very similar to (un)boundedness), and i'm in search of a precalculus solution.
 [/INST] As simple as it gets:
LEMMA: For $k\in N$ and $k>1$
$$\frac{2k}{2k-1}\gt\frac{\sqrt{3k+1}}{\sqrt{3k-2}}\tag{1}$$
Proof: It's trivial. Both sides are positive, square them:
$$\frac{4k^2}{4k^2-4k+1}\gt\frac{3k+1}{3k-2}$$
..which is equivalent to:
$$4k^2(3k-2)\gt(4k^2-4k+1)(3k+1)$$
$$12k^3-8k^2\gt12k^3+4k^2-12k^2-4k+3k+1$$
$$12k^3-8k^2\gt12k^3-8k^2-k+1$$
$$0\gt-k+1$$
...which is obviously true.
End of lemma proof.
By using the inequality (1) we have:
$$\frac{2}{1}=\frac{\sqrt{4}}{\sqrt{1}}$$
$$\frac{4}{3}\gt\frac{\sqrt{7}}{\sqrt{4}}$$
$$\frac{6}{5}\gt\frac{\sqrt{10}}{\sqrt{7}}$$
$$...$$
$$\frac{2n-2}{2n-3}\gt\frac{\sqrt{3n-2}}{\sqrt{3n-5}}\tag{1}$$
$$\frac{2n}{2n-1}\gt\frac{\sqrt{3n+1}}{\sqrt{3n-2}}\tag{1}$$
Multiply all this and you get:
$$a_n=\frac{(2n)!!}{(2n-1)!!} > \sqrt{3n+1}$$
So $a_n$ is obviously unbounded. Cute, isn't it? :)
</s>