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.

As far as monotonicity is concerned, write the sequence recursively:
\begin{align}
a_1 =&\  1 \\
a_{n} =&\ a_1\frac{n}{2n + 1} 
\end{align}
Then look at the difference between $a_{n + 1}$ and $a_{n}$:
\begin{align}
a_{n + 1} - a_{n} =&\ a_n\frac{n+1}{2(n + 1) + 1} - a_n\\
=&\ a_n\left(\frac{n+1}{2(n + 1) + 1} - 1\right) \\
=&\ a_n\left(\frac{n+1}{2n + 3} - 1\right) \\
=&\ a_n\frac{n + 1 - 2n - 3}{2n + 3} \\
=&\ a_n\frac{-n - 2}{2n + 3}\\
=&\ -a_n\frac{n + 2}{2n + 3}
\end{align}
Clearly this sequence monotonically decreases since the difference between each consecutive element is negative.
As far as convergence goes, one can certainly see that, in the limit that $n \rightarrow \infty$ that this difference will tend towards $\Delta a = -a_n\frac{1}{2}$.  The negative is showing that the next term is one half smaller than the previous.  For the sequence to converge, $\Delta a$ must tend towards $0$--the only possible way for this to happen is if $a_n$ eventually tends towards $0$.  Thus if the sequence converges, it converges to $0$.  And I put that in italics because I have not shown that the sequence converges--it seems to me that this is possible through the fact that it's positive definite and monotonically decreases (since it cannot monotonically decrease without crossing zero and not converge--but that's more of an intuitive argument).
