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.

$$|\frac{a_{n+1}}{a_n}|=|\frac{n+1}{2n+1}| \to\frac{1}{2}<1$$
So by the ratio test $\sum_{n=1}^{\infty} a_n$ converges and we must have $a_n \to 0$ by the divergence test.
In fact for $n \geq 1$, $\frac{n+1}{2n+1}$ is decreasing so,
$$\frac{a_{n+1}}{a_n}:=f(n) \leq \frac{1+1}{2(1)+1}=\frac{2}{3}$$
And because the positive reals are closed under multiplication it follows that,
$$0<a_{n+1} \leq \frac{2}{3}a_n$$
The solution to $\frac{a_{n+1}}{a_n}=f(n)$ is for $n >1$, $a_n=a_1 \prod_{x=1}^{n-1} f(x)$, but because $f(n)=:\frac{n+1}{2n+1} \leq \frac{2}{3}$ then for $n>1$:
$$0<a_n \leq a_1 \prod_{x=1}^{n-1} \frac{2}{3}=a_1 \left(\frac{2}{3}\right)^{n-1}$$
Because $x \geq a$ and $y \geq b$ $\implies$ $xy \geq ab$. And it's easy to show that the above inequality holds for $n=1$.
