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 $\lim_\limits{n\to\infty}a_n$ ?

We have
$$
\frac{\prod_{1}^{n}(2k-1)}{(2n)^{n}} < \frac{2n-1}{(2n)^{2}} \to 0
$$
as $n$ grows.
