Convergence and Limit of a Sequence I am looking at the sequence and I'm trying to see what happens as $n\to\infty$
$$
a_n = \frac{\prod_{1}^{n}(2n-1)}{(2n)^n} = \frac{1\cdot3\cdot5\cdot...\cdot(2n-1)}{(2n)^n}
$$
By inspection, I can see that the denominator increases faster than numerator which indicates that the sequence converges, but I'm not sure how to show that mathematically.
I tried using the ratio test to test for convergence but I got stuck on the following last step
$$
\begin{align*}
\lim_{n\to\infty}\left| \frac{a_{n+1}}{a_n}\right| &= \lim_{n\to\infty}\left|\frac{1\cdot3\cdot5\,\cdot\,...\,\cdot\,(2n-1)(2n+1)}{(2(n+1))^n}\cdot \frac{(2n)^n}{1\cdot3\cdot5\,\cdot\,...\,\cdot\,(2n-1)}\right| \\ &= \lim_{n\to\infty}\frac{n^n(2n+1)}{(n+1)^n} \\ &=\exp\left[\lim_{n\to\infty}\ln\left(\frac{n^n(2n+1)}{(n+1)^n}\right)\right]\\ &=\exp\left[\lim_{n\to\infty}n\ln \left(\frac{n}{n+1}\right) + \ln(2n+1)\right]
\end{align*}
$$
In short, how do I prove that this sequence is convergent, and how can I evaluate $\lim_\limits{n\to\infty}a_n$ ?

I am not sure that this is an answer to the question but, in any manner, it is too long for a comment.
Inserting the even numbers $$\prod_{i=1}^n (2i-1)=1 \cdot 3\cdot 5 \cdots (2n-1)=\frac {1 \cdot 2 \cdot 3 \cdot 4\cdots (2n-1)\cdot(2n)}{2\cdot 4\cdot 6\cdot 8\cdots \cdot(2n-2)\cdot(2n) }=\frac{(2n)!}{2^n n!}$$ This makes $$a_n = \frac{\prod_{i=1}^{n}(2i-1)}{(2n)^n} = \frac{(2n)!}{2^n\,(2n)^n \,n!}$$ Now, using Stirling approximation of the factorial $$m!\approx \sqrt{2 \pi\, m}\,m^m \,e^{-m}$$ after simplifications $$a_n\approx \sqrt{2} e^{-n}$$ which is a very good approximation and makes the problem much easier (I hope).
For illustration purposes $a_{10}=\frac{26189163}{409600000000}\approx 0.00006394$ while $\sqrt{2} e^{-10}\approx 0.00006421$.
Edit
If you use  Stirling's formula to two orders, $$m!\approx \sqrt{2 \pi m}\,m^{m}\,e^{-m} \left(1+\frac{1}{12 m}\right) $$ and use the same procedure as above, you should ned with $$a_n\approx \sqrt{2} e^{-n}\frac{ 1+\frac{1}{24 n}}{1+\frac{1}{12 n}}$$ whic, for $n=10$ would give $\approx 0.0000639399$ while the exact value would be $\approx 0.0000639384$
