Using squeeze thorem find $ \lim_{n \to \infty}{\frac{1 \cdot 3 \cdot 5 \cdot ...\cdot (2n-1)}{2\cdot 4 \cdot 6 \cdot ...\cdot 2n}}$ It is already solved here at Math.stackexchange, but we haven't learned Stirling's approximation (at our school), so can it be solved using only squeeze theorem?
$$\lim_{n \to \infty}{\frac{1 \cdot 3 \cdot 5 \cdot ...\cdot (2n-1)}{2\cdot 4 \cdot 6 \cdot ...\cdot 2n}}$$
My attempt,
Let $y_n = \frac{1 \cdot 3 \cdot 5 \cdot ...\cdot (2n-1)}{2\cdot 4 \cdot 6 \cdot ...\cdot 2n}$, we see that $ \frac{1 \cdot 3 \cdot 5 \cdot ...\cdot (2n-1)}{2\cdot 4 \cdot 6\cdot ...\cdot 2n} > \frac{1 \cdot 2 \cdot 2 \cdot ... \cdot 2}{2\cdot 4 \cdot 6 \cdot ...\cdot 2n} = \frac{1}{1 \cdot 2 \cdots .. \cdot (n-1)\cdot2n} = x_n$
$$\lim_{n \to \infty}{x_n} = 0 $$
Now I just need to find a sequnce $z_n>y_n$, so, $\lim_{n \to \infty} z_n = 0$. 

Let 
$$
b_n=\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots2n}.
$$
We shall show inductively that
$$
\frac{2}{\sqrt{2n+1}}>b_n.
$$
For $n=1$ it clearly holds.
Assume that
$$
\frac{2}{\sqrt{2k+1}}>\frac{1\cdot 3\cdots (2k-1)}{2\cdot 4\cdots2k}.
$$
Then
$$
\frac{2}{\sqrt{2k+1}}\cdot\frac{2k+1}{2k+2}>\frac{1\cdot 3\cdots (2k-1)\cdot(2k+1)}{2\cdot 4\cdots(2k)\cdot(2k+2)}.
$$
But
$$
\sqrt{\frac{2k+3}{2k+1}}=\sqrt{1+\frac{2}{2k+1}}<1+\frac{1}{2k+1}=\frac{2k+2}{2k+1}
$$
and hence
$$
\frac{2}{\sqrt{2k+3}}>\frac{2}{\sqrt{2k+1}}\cdot\frac{2k+1}{2k+2}>b_{k+1}.
$$
Another way to do it is by showing that
$$
a_n=\frac{2\cdots 4\cdots 2n}{1\cdots 3\cdots (2n-1)}\to\infty.
$$
Now, using the fact that,
$$
x\in[0,1]\Longrightarrow \log(1+x)=\int_0^x\frac{dt}{1+t}\ge \frac{1}{2}\int_0^x\frac{dt}{1+t}=\frac{x}{2},
$$
we obtain
$$
\log a_n=\sum_{k=1}^n\log \left(\frac{2k}{2k-1}\right)=\sum_{k=1}^n\log \left(1+\frac{1}{2k-1}\right)\ge \frac{1}{2}\sum_{k=1}^n \frac{1}{2k-1}.
$$
It only remains to show that
$$
1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n-1}\to\infty.
$$
