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.
 A: 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 $0$.
A: $$|\frac{a_{n+1}}{a_n}|=|\frac{n+1}{2n+1}| \to\frac{1}{2}<1$$
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 $n=1$.
A: 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 $A_n\to 0$.
A: You can use the odd-factorial
$$\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)} = \frac{(n!)^2 2^n}{(2n)!}
$$
Taking logarithm,
$$ \log\frac{(n!)^2 2^n}{(2n)!} = 2\log n! + n \log 2 - \log (2n!)
$$
so...
A: $a_{n+1} = a_n \frac{n+1}{2n+1}$  Each number in the sequence is slightly more that (1/2) as the one before it.
for $n>3, \dfrac{a_{n+1}}{a_n} < 0.6$
$a_n < 0.4 (0.6)^{n-2}$
The sequence converges to 0.
A: 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).
A: Just manipulate the Wallis product for $\pi$.
