Does the sequence $\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}$ converge?

The reciprocal of the term of interest is

$$\begin{align} \frac{(2n-1)!!}{n!}&=\left(\frac{(2n-1)}{n}\right)\left(\frac{(2(n-1)-1)}{(n-1)}\right)\left(\frac{(2(n-2)-1)}{(n-2)}\right) \cdots \left(\frac{5}{3}\right)\left(\frac{3}{2}\right)\\\\ &=\left(2-\frac{1}{n}\right)\left(2-\frac{1}{n-1}\right)\left(2-\frac{1}{n-2}\right) \cdots \left(\frac{5}{3}\right)\left(\frac{3}{2}\right)\\\\ &\ge \left(\frac32\right)^{n-1} \end{align}$$

Therefore, we see immediately that the limit of interest is $0$.

$$|\frac{a_{n+1}}{a_n}|=|\frac{n+1}{2n+1}| \to\frac{1}{2}<1$$

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 $n=1$.