Does the sequence $\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}$ converge? I'm trying to determine if this sequence converges as part of answering whether it's monotonic:
$$
\left\{\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}\right\}
$$
First, I tried expanding it a bit to see if I could remove common factors in the numerator and denominator:
$$
\left\{\frac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...\cdot n}{1\cdot 3\cdot 5\cdot 7\cdot 9 \cdot ...\cdot (2n-1)}\right\}
$$
Second, I tried looking at elements of the sequence with common factors removed:
$$
1, \frac{2}{3}, \frac{2}{5}, \frac{2\cdot 4}{5\cdot 7}, \frac{2\cdot 4}{7\cdot 9}, ...
$$
Third, I tried looking at the elements again as fractions without simplifications:
$$
\frac{1}{1}, \frac{2}{3}, \frac{6}{15}, \frac{24}{105}, \frac{120}{945}, ...
$$
Last, I tried searching for similar questions on Stack Exchange and I found one for $\frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)}$ but I didn't understand how that might apply to my question. So, any hints would be much appreciated.

Using Henry W.'s answer $$A_n=\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)} = 2^n\frac{(n!)^2 }{(2n)!}$$ $$\log(A_n)= 2\log(n!) + n \log( 2) - \log \big((2n)!\big)$$ Now, using Stirling approximation
$$\log(p!)\approx p\log(p)-p+\frac 12\log(2\pi p)$$ $$\log(A_n)\approx \frac{1}{2} \log (\pi  n)-n \log (2)$$ $n$ varies faster than $\log(n)$; so $\log(A_n)\to -\infty$ and $A_n\to 0$.
