Im tempted to say that the limit of this sequence is 1 because infinite root of infinite number is close to 1 but maybe Im mising here something? What will be inside the root?
This is the sequence:
$$\lim_{n\to\infty}\sqrt[n]{\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}}$$
Note that $$1\ge\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}=\frac{3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot(2n-2)}\cdot\frac1{2n}\ge\frac1{2n}$$ So the limit is $1$ by squeezing as $\sqrt[n]n\to1.$
Note that $$ \left[\frac1{2n+1}\right]^{1/n}=\left[\frac13\frac35\frac57\cdots\frac{2n-1}{2n+1}\right]^{1/n}\le\left[\frac12\frac34\frac56\cdots\frac{2n-1}{2n}\right]^{1/n}\lt1 $$ Since $\lim\limits_{n\to\infty}(2n+1)^{1/n}=1$, the Squeeze Theorem says that $$ \lim_{n\to\infty}\left[\frac12\frac34\frac56\cdots\frac{2n-1}{2n}\right]^{1/n}=1 $$
Addendum
Since $1+x\le e^x$ for all $x\in\mathbb{R}$, we easily have $$ (1+\sqrt{n}/2)^2\le \left(e^{\sqrt{n}/2}\right)^2 $$ which implies that $$ 1+2n\le8e^{\sqrt{n}} $$ Therefore $$ 1\le(1+2n)^{1/n}\le8^{1/n}e^{1/\sqrt{n}} $$ By the Squeeze Theorem, $$ \lim_{n\to\infty}(2n+1)^{1/n}=1 $$
If you insert the denominator into the numerator you have $$\frac{1\times 3\times 5\cdots \times(2n-1)}{2\times 4\times 6\cdots \times(2n)}=\frac{(2n)!}{4^n(n!)^2}$$ Now, use Stirling approximation $$m!=m^m \sqrt{2\pi m}e^{-m}$$
It is quite easy to show by induction that: $$\frac{(2n-1)!!}{(2n)!!}\geq\frac{1}{\sqrt{2n}}\tag{1}$$ since the last line is implied by: $$\frac{2n+1}{2n+2}\geq\sqrt{\frac{2n}{2n+2}}$$ that is equivalent to: $$(2n+1)^2\geq 2n(2n+2) = (2n+1)^2-1.$$ Using $(1)$ and the trivial bound $\frac{(2n-1)!!}{(2n)!!}\leq 1$, it follows that the limit is $1$ by squeezing.
You may write $$ \begin{align} \frac{1\cdot 3\cdot 5 \cdots(2n-1)}{2\cdot 4\cdot 6\cdots2n} &=\frac{1\cdot 2\cdot 3\cdot 4 \cdot 5\cdot6\cdots(2n-1)\cdot 2n}{(2\cdot 4\cdot 6\cdots (2n))^2}\\ &=\frac{(2n)!}{(2^{n} \cdot 1\cdot 2\cdot 3 \cdot 4 \cdots n)^2}\\ & =\frac{(2n)!}{2^{2n} (n!)^2 }\\ & =\frac{1}{\sqrt{\pi n}}+\mathcal{O}\left(\frac{1}{n^{3/2}}\right), \quad \text{for} \, n \, \text{great} \end{align} $$ where we have use Stirling's approximation, then you easily conclude, since $$ \sqrt[n]{\frac{1\cdot 3\cdot 5 \cdots(2n-1)}{2\cdot 4\cdot 6\cdots2n}}=e^{\Large \frac{1}{n}\log{\frac{1\cdot 3\cdot 5 \cdots(2n-1)}{2\cdot 4\cdot 6\cdots2n}}} $$ giving the value $1$ for your limit.
Taking log and Cesaro-Stolz: $$\eqalign{\log L &=\lim_{n\to\infty}\frac{(\log(1)+\log(3)+\ldots+\log(2n-1))-(\log(2)+\log(4)+\ldots+\log(2n))}n=\cr &=\lim_{n\to\infty}\frac{\log(2n+1)-\log(2n+2)}1=0.}$$
A recent question here showed that the expression inside the root was roughly $\sqrt{1/n}$. The $n$th root of that approaches 1.
$$\log f(n)=\frac1n\left(\log\frac12+\log\frac34+...+\log\frac{2n-1}{2n}\right)\\
\approx\frac1n\left(C_1-\frac12-\frac14\cdots-\frac1{2n}\right)\\
\approx\frac1n\left(C_2-\log n\right)/2\\
\to0\text{ as }n\to\infty$$
Hint. Multiply numerator and denominator by $(2 \cdot 4 \cdots 2n)$ and remember that $2 \cdot 4 \cdots 2n = 2^n \cdot n!$
If you multiply top and bottom by $2\cdot 4\cdot 6\cdots 2n$ and simplify, you should get an expression $\sqrt[n]{\frac{(2n)!}{4^n(n!)^2}}$, which can be approximated using Stirling's formula.
Using the following inequality \begin{equation} \frac{n}{\sum_{k=1}^{n} \frac{1}{1-\frac{1}{2k}} } \leq \sqrt[n]{(1-\frac{1}{2})(1-\frac{1}{4})...(1-\frac{1}{2n})} \leq \frac{(1-\frac{1}{2})+(1-\frac{1}{4})+...,(1-\frac{1}{2n})}{n} = 1-\frac{1}{2n}\sum_{k=1}^{n}\frac{1}{k} \end{equation} and another results \begin{equation} \sum_{k=1}^{n}\frac{1}{k} = \ln(n) + \gamma ,n \rightarrow \infty, \end{equation} where $\gamma $ is Euler constant. and \begin{equation} \sum_{k=1}^{n} \frac{1}{1-\frac{1}{2k}}=\sum_{k=1}^{n}\frac{2k-1 +1}{2k-1} = n + \sum_{k=1}^{n}\frac{1}{2k-1}\leq n + \sum_{k=1}^{n}\frac{1}{k} \end{equation}
Hence we have \begin{equation} \frac{n}{n+\ln(n) + \gamma} \leq \sqrt[n]{(1-\frac{1}{2})(1-\frac{1}{4})...(1-\frac{1}{2n})} \leq 1 - \frac{\ln(n) + \gamma}{2n} \end{equation}
Let $$a _n=\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}.$$ Note that $$ \frac{a_{n+1}}{a_n}=\frac{2n+1}{2n+2}\to1 $$ as $n\to \infty$. By this MSE post which proves that, $$ \liminf(b_{n+1}/b_n) \leq \liminf(b_n^{1/n}) \leq \limsup(b_n^{1/n}) \leq \limsup(b_{n+1}/b_n) $$ for any sequence $(b_n)$ such that $b_n>0$, it follows that $$ a_n^{1/n}\to1 $$ as $n\to\infty$.
